DEV Community: fengmao

史上最全的文档地址列表

fengmao — Sun, 22 May 2022 08:42:00 +0000

周末花了两天时间收集了几百个网站的文档地址，类型从前端到后端全部覆盖了，还有一些热门的技术，也添加进来了，这里分享给大家，统一的入口地址在adochub.

docker
nginx
python3
meteor
prisma
vuejs
anaconda
caddy server
divio
aws
mongoosejs
tidb
flutter
gitlab
threejs
microsoft
pytorch
moodle
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github
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nodejs
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npmjs
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latex project
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love
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akka
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clickhouse

step by step guide tell you how to build a website like apkmirror

fengmao — Mon, 16 Aug 2021 12:58:48 +0000

There are many free apk download websites such as apkmirror, today i will tell you how to build a website like apkmirror, the programming language i used is node.js, the database i used is mongodb, search engine used is elasticsearch, the web framework i use is eggjs.

Part I analysis apk file and save data

The tool we need is appt, you can install it with sudo apt-get install aapt on ubuntu, the command to analysis apk file is aapt dump badging test.apk, as we need execute the command in node.js script so i use execSync to call the command, the code block is below.

const { execSync } = require('child_process');
const fs = require("fs"); 
const md5 = require('md5');
const { sleep, getDir } = require('../util');

const analysis = (output) => {
  const res = {};
  let out = output.toString().replace(/'/g, '');

  out = out.split('\n');

  let tmp = out[0].split(' ');
  res.package = tmp[1].split('=')[1];
  res.versionCode = tmp[2].split('=')[1];
  res.versionName = tmp[3].split('=')[1];

  tmp = out[2].split(':')
  if (tmp[0] == 'sdkVersion') {
    res.sdkVersion = tmp[1];
  }

  tmp = out[3].split(':');
  if (tmp[0] == 'targetSdkVersion') {
    res.targetSdkVersion = tmp[1];
  }

  tmp = out[out.length - 2].split(':');
  if (tmp[0] == 'native-code') {
    res.variant = tmp[1].trim().replace(/ /g, '|');

    tmp = out[out.length - 6].split(':');
    res.screen = tmp[1].trim();

    tmp = out[out.length - 4].split(':');
    res.locales = tmp[1].trim();
  } else {
    res.variant = 'noarch';

    tmp = out[out.length - 5].split(':');
    res.screen = tmp[1].trim();

    tmp = out[out.length - 3].split(':');
    res.locales = tmp[1].trim();
  }
  return res;
}

const filters = ['.', '..'];

const run = async () => {
  while (true) {
    const dirs = fs.readdirSync(getDir('tmp'));
    for (const dir of dirs) {
      if (!filters.includes(dir)) {
        const name = getDir('tmp', dir);
        try {
          const output = execSync(`/usr/bin/aapt dump badging ${name}`);
          const res = analysis(output);
          const stats = fs.statSync(name);
          res.fileSize = stats.size;
          const newName = getDir('apk', `${res.package}_${res.versionCode}_idoras.com.apk`);
          const tasks = [];
          const url = `https://play.google.com/store/apps/details?id=${res.package}`;
          tasks.push({ _id: md5(url), url, host: 'play.google.com', type: 'detail', package: res.package, done: 0 });
          await global.db.collection('tasks').insertMany(tasks, { ordered: false }).catch(() => {});
          const one = await global.db.collection('apk').findOne({ _id: res.package }).catch(() => {});
          if (one) {
            const a = +one.versionCode;
            const b = +res.versionCode;
            if (b > a) {
              await global.db.collection('apk').updateOne({ _id: res.package }, { $set:{ udate: Date.now(), versionCode: res.versionCode, versionName: res.versionName, done: 1 }}).catch(() => {});
            }
          } else {
            await global.db.collection('apk').insertOne({ _id: res.package, versionCode: res.versionCode, versionName: res.versionName, cdate: Date.now(), udate: Date.now(), done: 0, down: 0 }).catch(() => {});
          }
          res._id = `${res.package}_${res.versionCode}`;
          await global.db.collection('variant').insertOne({ ...res, cdate: Date.now() }).catch(() => {});
          fs.renameSync(name, newName);
        } catch (err) {
          console.error(err);
          fs.unlinkSync(name);
        }
      }
    }
    if (dirs.length == 0) {
      console.log('no apk files to analysis');
      await sleep(60000);
    }
  }
};

module.exports = run;

The getDir and sleep is util function, you can find the source code below.

const req = require('request');
const path = require('path');
const os = require('os');
const agent = new require('socks-proxy-agent')('socks5://localhost:1080');

let root = '/data';
const options = {
  headers: {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/92.0.4515.107 Safari/537.36'
  },
  timeout: 30000,
  encoding: null,
};
if (os.platform() == 'win32') {
  options.agent = agent;
  root = 'd:/data';
}
const request = req.defaults(options);

const getDir = (...paths) => {
  paths.unshift(root);
  return path.join(...paths);
};

const sleep = (time) => new Promise((resolve) => {
    setTimeout(() => {
      resolve();
    }, time);
  });

const fetch = (url) => new Promise((resolve) => {
    console.log(`down ${url} started`);
    request(url, (err, res, body) => {
      if (res && res.statusCode === 200) {
        console.log(`down ${url} 200`);
        resolve(body);
      } else {
        console.error(`down ${url} ${res && res.statusCode} ${err}`);
        if (res && res.statusCode) {
          resolve(res.statusCode);
        } else {
          // ESOCKETTIMEOUT 超时错误返回600
          resolve(600);
        }
      }
    });
  });

module.exports = { getDir, sleep, fetch };

The main logic is we get basic information from appt, parse the output, get apk id such as com.moonvideo.android.resso, and insert a task to get apk logo and images and brief from play.google.com, i wrote a little crawler framework, each website with one parser bind with the host, every url have a type, each type bind with a function, the play.google.com’s parser code is below.

const { URL } = require('url');
const md5 = require('md5');
const cheerio = require('cheerio');
const slugify = require('slugify');

const resolve = (from, to) => {
  const resolvedUrl = new URL(to, new URL(from, 'resolve://'));
  if (resolvedUrl.protocol === 'resolve:') {
    // `from` is a relative URL.
    const { pathname, search, hash } = resolvedUrl;
    return pathname + search + hash;
  }
  return resolvedUrl.toString();
}

const fns = {

  detail: async (page) => {
    const host = page.host;
    try {
      const $ = cheerio.load(page.con.toString(), { decodeEntities: false });
      const brief = $('div.DWPxHb > span > div:nth-child(1)').html();
      const name = $('h1.AHFaub > span').text();
      const dev = $('div.qQKdcc > span:nth-child(1) > a').text();
      const tag = $('div.qQKdcc > span:nth-child(2) > a').attr('href').split('/')[4];
      const images = [];
      const tasks = [];
      $('button.Q4vdJd > img').each((i, ele) => {
        let url = $(ele).attr('src');
        if (!url || !url.startsWith('https')) {
          url = $(ele).attr('data-src')
        }
        tasks.push({ _id: md5(url), url, type: 'download', host: (new URL(url)).hostname, done: 0 });
        images.push({ _id: md5(url), url });
      });
      const logourl = $('div.xSyT2c > img').attr('src');
      tasks.push({ _id: md5(logourl), url: logourl, type: 'download', host: (new URL(logourl)).hostname, package: page.package, done: 0 });
      const res = await global.db.collection('tasks').insertMany(tasks, { ordered: false }).catch(() => {});
      res && console.log(`${host}-detail insert ${res.insertedCount} from ${tasks.length} tasks`);
      await global.db.collection('apk').updateOne({ _id: page.package }, { $set:{ brief, name, dev, devSlug: slugify(dev), tag, tagSlug: slugify(tag), images, done: 1 } });
      return 1;
    } catch (err) {
      console.error(`${host}-detail parse ${page.url} ${err}`);
      return 4;
    }
  },

  run: (page) => {
    const fn = fns[page.type];
    if (fn) {
      return fn(page);
    }
    console.error(`${page.url} parser not found`);
    return 0;
  }

};

module.exports = fns;

every apk we analysis with aapt will send one request to play.google.com to get it’s name, brief, developer info, tag, images, i will use these data to show on the website.

now we have the data of apk file, so where to find the apk files? You can download them from idoras or apkmirror or apkpure, also you can write node.js script to download them from website such as idoras.

Next we need index these data with elasticsearch so we can use them later, the code is below.

const { sleep } = require('../util');
const { Client } = require('@elastic/elasticsearch');

const getApks = (apks) => {
  const res = [];
  apks.forEach((apk) => {
    if(apk.name) {
      const tmp = {};
      res.push({ index: { _index: 'apk', _type: '_doc', _id: apk._id } });
      tmp.id = apk._id;
      tmp.versionCode = apk.versionCode;
      tmp.versionName = apk.versionName;
      tmp.cdate = apk.cdate;
      tmp.udate = apk.udate;
      tmp.down = apk.down;
      tmp.brief = apk.brief;
      tmp.dev = apk.dev;
      tmp.devSlug = apk.devSlug;
      tmp.name = apk.name;
      tmp.tag = apk.tag;
      res.push(tmp);
    }
  });
  return res;
};

const run = async () => {

  while (true) {
    const esclient = new Client({
      node: 'http://username:password@ip:9200',
      maxRetries: 3,
      requestTimeout: 60000
    });
    const apks = await global.db.collection('apk').find({ done: 1 }).limit(100).toArray().catch((err) => console.error(err));
    const tmp = getApks(apks);
    if (tmp && tmp.length) {
      let res = await esclient.bulk({ refresh: true, body: tmp }).catch((err) => console.error(err));
      if (res) {
        const ulist = apks.map((apk) => ({ updateOne: { filter: { _id: apk._id }, update: { $set: { done: 6 } }, upsert: false } }));
        res = await global.db.collection('apk').bulkWrite(ulist, { ordered: false, w: 1 }).catch(() => {});
        res && console.log(`apk:indexer update ${res.modifiedCount} from ${res.matchedCount}`);
      }
    } else {
      console.log('apk:indexer no apk to index');
      await sleep(60000);
    }
    await esclient.close();
  }
};

module.exports = run;

Part II build website with eggjs

The template engine i use is nunjucks, also i used some eggjs plugins such egg-elasticsearch2

The logic is easy, at home page we query data from elasticsearch, and render data into nunjucks template, basiclly it’s a list page, also i will cache the hot download apks into ctx.app.hots, refresh it every 30 minutes, and show the hot download apks on the right aside, also the html template support responsive design, it will show perfect on mobile system.

'use strict';

const Controller = require('egg').Controller;

class HomeController extends Controller {

  async home() {
    const { ctx } = this;
    const params = ctx.params;
    const page = parseInt(params.page, 10) || 1;
    if (page < 1) {
      ctx.redirect('/list/1/', 301);
    }
    const start = (page - 1) * ctx.helper.size;
    const query = {
      body: [
        { index: 'apk', type: '_doc' },
        { query: { bool: { must: [{ match_all: {} }, { exists: { field: 'id' } }] } }, highlight: { fields: { name: {} } }, from: start, size: ctx.helper.size, sort: [{ udate: { order: 'desc' } }, '_score' ] },
      ],
    };
    const info = {};
    const res = await ctx.app.elasticsearch.msearch(query);
    ctx.helper.getList(res.responses[0], info);

    let tpage = Math.floor(info.all / ctx.helper.size) + (info.all % ctx.helper.size ? 1 : 0);
    tpage < 1 && (tpage = 1);
    if (page > tpage) {
      if (tpage > 500) {
        ctx.redirect('/list/500/', 301);
      } else {
        ctx.redirect(`/list/${tpage}/`, 301);
      }
    } else if (page > 500) {
      ctx.redirect('/list/500/', 301);
    }
    tpage = tpage > 500 ? 500 : tpage;
    info.pages = ctx.helper.getPages(page, tpage);

    info.url = ctx.request.url;
    if (info.url === '/') {
      info.title = 'Free Apk download online - idoras.com';
      info.keywords = 'app download,apk download, free apk download, idoras, apk downloader, android apk download';
      info.description = 'Free apk download for Android with idoras APK downloader. NoAds, Faster apk downloads and apk file update speed. Best of all, it\'s free';
      info.home = true;
    } else {
      info.title = `Free Apk download - page${page} - idoras.com`;
      info.keywords = `app download,apk download, free apk download, idoras, apk downloader, android apk download,apk list page${page}`;
      info.description = info.list.map(item => item.name).join(',').substring(0, 150);
    }

    if (!ctx.app.hots) {
      ctx.app.hots = info.hots = await ctx.service.search.hot();
    } else {
      info.hots = ctx.app.hots;
    }
    info.slist = ctx.helper.getKeys();
    await ctx.render('home.html', { info });
  }

  async version() {
    const { ctx } = this;
    const params = ctx.params;
    const id = params.id;
    const apk = await ctx.model.Apk.findById(id);
    const list = await ctx.model.Variant.find({ package: id });
    list.sort((a, b) => b.versionCode - a.versionCode);
    const info = { apk, list };
    info.variant = list[0] || {};
    info.title = `${apk.name} Free Apk download - idoras.com`;
    info.keywords = `${apk.name} download, app download,apk download, free apk download, idoras, apk downloader, android apk download`;
    info.description = `download ${apk.name} free apks online, it's faster and free online, many version to download`;
    if (!ctx.app.hots) {
      ctx.app.hots = info.hots = await ctx.service.search.hot();
    } else {
      info.hots = ctx.app.hots;
    }
    info.slist = ctx.helper.getKeys();
    await ctx.render('version.html', { info });
  }

  async download() {
    const { ctx } = this;
    const params = ctx.params;
    const id = params.id;
    const version = params.version;
    const apk = await ctx.model.Apk.findById(id);
    const variant = await ctx.model.Variant.findById(`${id}_${version}`);
    const list = await ctx.model.Variant.find({ package: id });
    list.sort((a, b) => b.cdate - a.cdate);
    apk.down = apk.down + 1;
    apk.save();
    ctx.app.elasticsearch.update({ index: 'apk', type: '_doc', id, body: { script: 'ctx._source.down += 1', upsert: { down: 1 } } }, () => { });
    const info = { apk, variant, list };
    info.title = `${apk.name} ${apk.versionName} Free Apk download online - idoras.com`;
    info.keywords = `${apk.name} ${apk.versionName} download online, app download,apk download, free apk download, idoras, apk downloader, android apk download`;
    info.description = `download ${apk.name} ${apk.versionName} free apks online, it's faster and free online, many version to download`;
    if (!ctx.app.hots) {
      ctx.app.hots = info.hots = await ctx.service.search.hot();
    } else {
      info.hots = ctx.app.hots;
    }
    info.slist = ctx.helper.getKeys();
    await ctx.render('download.html', { info });
  }

  async dev() {
    const { ctx } = this;
    const params = ctx.params;
    const devSlug = params.devSlug;
    const page = parseInt(params.page, 10) || 1;
    if (page < 1) {
      ctx.redirect(`/dev/${devSlug}/1/`, 301);
    }
    const start = (page - 1) * ctx.helper.size;
    const info = { devSlug };
    const query = {
      body: [
        { index: 'apk', type: '_doc' },
        { query: { term: { devSlug } }, from: start, size: ctx.helper.size, sort: [{ udate: { order: 'desc' } }, '_score' ] },
      ],
    };
    const res = await ctx.app.elasticsearch.msearch(query);
    ctx.helper.getList(res.responses[0], info);

    let tpage = Math.floor(info.all / ctx.helper.size) + (info.all % ctx.helper.size ? 1 : 0);
    tpage < 1 && (tpage = 1);
    if (page > tpage) {
      if (tpage > 500) {
        ctx.redirect(`/dev/${devSlug}/500/`, 301);
      } else {
        ctx.redirect(`/dev/${devSlug}/${tpage}/`, 301);
      }
    } else if (page > 500) {
      ctx.redirect(`/dev/${devSlug}/500/`, 301);
    }
    tpage = tpage > 500 ? 500 : tpage;
    info.pages = ctx.helper.getPages(page, tpage);

    info.title = `apk developered by ${devSlug} download - idoras.com`;
    info.keywords = `${devSlug} apks download online, app download,apk download, free apk download, idoras, apk downloader, android apk download`;
    info.description = `download ${devSlug} free apks online, it's faster and free online, many version to download`;
    if (!ctx.app.hots) {
      ctx.app.hots = info.hots = await ctx.service.search.hot();
    } else {
      info.hots = ctx.app.hots;
    }
    info.slist = ctx.helper.getKeys();
    await ctx.render('dev.html', { info });
  }

  async tag() {
    const { ctx } = this;
    const params = ctx.params;
    const tag = params.tag;
    const page = parseInt(params.page, 10) || 1;
    if (page < 1) {
      ctx.redirect(`/tag/${tag}/1/`, 301);
    }
    const start = (page - 1) * ctx.helper.size;
    const info = { tag };
    const query = {
      body: [
        { index: 'apk', type: '_doc' },
        { query: { term: { tag } }, from: start, size: ctx.helper.size, sort: [{ udate: { order: 'desc' } }, '_score' ] },
      ],
    };
    const res = await ctx.app.elasticsearch.msearch(query);
    ctx.helper.getList(res.responses[0], info);

    let tpage = Math.floor(info.all / ctx.helper.size) + (info.all % ctx.helper.size ? 1 : 0);
    tpage < 1 && (tpage = 1);
    if (page > tpage) {
      if (tpage > 500) {
        ctx.redirect(`/tag/${tag}/500/`, 301);
      } else {
        ctx.redirect(`/tag/${tag}/${tpage}/`, 301);
      }
    } else if (page > 500) {
      ctx.redirect(`/tag/${tag}/500/`, 301);
    }
    tpage = tpage > 500 ? 500 : tpage;
    info.pages = ctx.helper.getPages(page, tpage);

    info.title = `apk tagged with ${tag} download - idoras.com`;
    info.keywords = `${tag} apks download online, app download,apk download, free apk download, idoras, apk downloader, android apk download`;
    info.description = `download ${tag} free apks online, it's faster and free online, many version to download`;
    if (!ctx.app.hots) {
      ctx.app.hots = info.hots = await ctx.service.search.hot();
    } else {
      info.hots = ctx.app.hots;
    }
    info.slist = ctx.helper.getKeys();
    await ctx.render('tag.html', { info });
  }

  async search() {
    const { ctx } = this;
    const params = ctx.params;
    const key = params.key;
    const page = parseInt(params.page, 10) || 1;
    if (page < 1) {
      ctx.redirect(`/search/${key}/1/`, 301);
    }
    const start = (page - 1) * ctx.helper.size;
    const info = { key };
    const query = {
      body: [
        { index: 'apk', type: '_doc' },
      ],
    };
    if (key) {
      query.body.push({ query: { multi_match: { query: key, fields: [ 'name', 'id', 'brief' ] } }, from: start, size: ctx.helper.size, sort: [ '_score' ] });
    } else {
      query.body.push({ query: { bool: { must: [{ match_all: {} }, { exists: { field: 'id' } }] } }, from: start, size: ctx.helper.size, sort: [ '_score' ] });
    }
    const res = await ctx.app.elasticsearch.msearch(query);
    key && res.responses[0].hits.hits.length && ctx.helper.putKey(key);
    ctx.helper.getList(res.responses[0], info);

    let tpage = Math.floor(info.all / ctx.helper.size) + (info.all % ctx.helper.size ? 1 : 0);
    tpage < 1 && (tpage = 1);
    if (page > tpage) {
      if (tpage > 500) {
        ctx.redirect(`/search/${key}/500/`, 301);
      } else {
        ctx.redirect(`/search/${key}/${tpage}/`, 301);
      }
    } else if (page > 500) {
      ctx.redirect(`/search/${key}/500/`, 301);
    }
    tpage = tpage > 500 ? 500 : tpage;
    info.pages = ctx.helper.getPages(page, tpage);

    info.title = `search ${key} apks to download online - idoras.com`;
    info.keywords = `${key} apks download online, app download,apk download, free apk download, idoras, apk downloader, android apk download`;
    info.description = `download ${key} free apks online, it's faster and free online, many version to download`;
    if (!ctx.app.hots) {
      ctx.app.hots = info.hots = await ctx.service.search.hot();
    } else {
      info.hots = ctx.app.hots;
    }
    info.slist = ctx.helper.getKeys();
    await ctx.render('search.html', { info });
  }

  async re() {
    const { ctx } = this;
    const params = ctx.request.body;
    const key = params.key;
    if (key) {
      ctx.redirect(`/search/${encodeURI(key)}/`, 301);
    } else {
      ctx.redirect('/search/', 301);
    }
  }
}

module.exports = HomeController;

'use strict';

// eslint-disable-next-line no-unused-vars
module.exports = (options, app) => {
  return async function mobileMiddleware(ctx, next) {
    const u = ctx.get('user-agent') || '';
    const tmp = {
      trident: u.indexOf('Trident') > -1, // IE内核
      presto: u.indexOf('Presto') > -1, // opera内核
      webKit: u.indexOf('AppleWebKit') > -1, // 苹果、谷歌内核
      gecko: u.indexOf('Gecko') > -1 && u.indexOf('KHTML') === -1, // 火狐内核
      mobile: !!u.match(/AppleWebKit.*Mobile.*/), // 是否为移动终端
      ios: !!u.match(/\(i[^;]+;( U;)? CPU.+Mac OS X/), // ios终端
      android: u.indexOf('Android') > -1 || u.indexOf('Linux') > -1, // android终端或者uc浏览器
      iPhone: u.indexOf('iPhone') > -1, // 是否为iPhone或者QQHD浏览器
      iPad: u.indexOf('iPad') > -1, // 是否iPad
      webApp: u.indexOf('Safari') === -1, // 是否web应该程序，没有头部与底部
      weixin: u.indexOf('MicroMessenger') > -1, // 是否微信 （2015-01-22新增）
      qq: u.match(/\sQQ/i) === ' qq', // 是否QQ
    };
    if (tmp.mobile || tmp.android || tmp.ios || tmp.weixin) {
      ctx.request.mobile = true;
    }
    await next();
  };
};

<!DOCTYPE html>
<html lang="en">

<head>
  <meta charset="UTF-8">
  <meta http-equiv="X-UA-Compatible" content="IE=edge">
  <meta name="viewport" content="width=device-width, initial-scale=1.0">
  <title>{{info.title}}</title>
  <meta name="keywords" content="{{info.keywords}}">
  <meta name="description" content="{{info.description}}">
  <meta property="og:site_name" content="idoras.com">
  <meta property="og:title" content="{{info.title}}">
  <meta property="og:type" content="website">
  <link rel="shortcut icon" href="/favicon.ico">
  <link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/normalize/8.0.0/normalize.min.css">
  <link rel="stylesheet" href="https://fonts.googleapis.com/icon?family=Material+Icons">
  <link rel="stylesheet" href="/css/material-components-web.min.css">
  <link rel="stylesheet" href="/css/app.css?v=2021080817">
</head>

<body>

  <aside class="mdc-drawer mdc-drawer--dismissible">
    <div class="mdc-drawer__header">
      <h3 class="mdc-drawer__title">IDORAS</h3>
      <h6 class="mdc-drawer__subtitle">download free android apks</h6>
    </div>
    <div class="mdc-drawer__content">
      <nav class="mdc-list">
        <a class="mdc-list-item mdc-list-item--activated" href="/" aria-selected="true">
          <i class="material-icons mdc-list-item__graphic" aria-hidden="true">home</i>
          <span class="mdc-list-item__text">Home</span>
        </a>
      </nav>
    </div>
  </aside>
  <div class="mdc-drawer-app-content">
    <header class="mdc-top-app-bar app-bar" id="app-bar">
      <div class="mdc-top-app-bar__row container">
        <section class="mdc-top-app-bar__section mdc-top-app-bar__section--align-start">
          <i href="" class="demo-menu material-icons mdc-top-app-bar__navigation-icon mdc-nav-icon menuLogo"
            style="text-decoration: none;cursor: pointer;">menu</i>
          <a class="mdc-top-app-bar__title" href="/" style="color: #fff;">IDORAS</a>
        </section>
        <section class="mdc-top-app-bar__section mdc-top-app-bar__section--align-end" role="toolbar">
          <a href="/search/">
            <i class="material-icons icon-white">search</i>
          </a>
        </section>
      </div>
    </header>

    <main class="main-content" id="main-content">
      <div class="mdc-top-app-bar--fixed-adjust container">
        <div class="mdc-layout-grid" style="padding:0">
          <div class="mdc-layout-grid__inner">
            <div
              class="mdc-layout-grid__cell mdc-layout-grid__cell--span-8 mdc-layout-grid__cell--span-12-tablet mdc-layout-grid__cell--span-12-phone">
              {% if ctx.request.mobile %}
              <div class="details-title">
                <!-- Go to www.addthis.com/dashboard to customize your tools --> 
                <div class="addthis_inline_share_toolbox_bm88"></div>
              </div>
              {% endif %}
              <div class="listWidght">
                <form action="/re" style="overflow:visible" method="POST" onsubmit="return key.value!=''" role="search">
                  <input class="search-key" type="text" id="key" name="key" placeholder="Google Play Store or com.android.vending" size="1" style="padding: 5px;" />
                  <button type="submit" class="search mdc-button mdc-button--raised button">
                    <i class="material-icons icon-white">search</i>
                  </button>
                </form>
              </div>
              <div class="listWidget">
                <div class="box">
                  <div class="title bread-crumbs">
                    <h5 class="widgetHeader">Latest Upload</h5>
                  </div>
                </div>
                {% for item in info.list %}
                <div>
                  <div class="appRow">
                    <div class="table-row">
                      <div style="width: 56px;" class="table-cell">
                        <img style="width:50px; height:50px;"
                          src="https://img.idoras.com/img/{{item.id}}.png">
                      </div>
                      <div class="table-cell">
                        <div style="padding-top: 4px;">
                          <h5 title="{{item.name}} {{item.versionName}} ({{item.versionCode}}) by {{item.dev}}"
                            class="appRowTitle wrapText marginZero block-on-mobile">
                            <a class="fontBlack" href="/apk/{{item.id}}/">{{item.name}} {{item.versionName}}</a>
                          </h5>
                          <a href="/dev/{{item.devSlug}}/" class="byDeveloper block-on-mobile wrapText">by {{item.dev}}</a>
                          <span style="padding:4px 12px 0 0;" class="colorLightBlack"><span class="udate">{{ctx.helper.fdate(item.udate)}}</span></span>
                        </div>
                      </div>
                      <div style="width: 30px;" class="table-cell">
                        <div class="iconsBox ">
                          <div class="downloadIconPositioning">
                            <a class="downloadLink" href="/apk/{{item.id}}/{{item.versionCode}}/">
                              <i class="material-icons icon-blue" aria-hidden="true">download</i>
                            </a>
                          </div>
                        </div>
                      </div>
                    </div>
                  </div>
                </div>
                {% endfor %}
                <div class="appRow center">
                  <div class="pagination mobile">
                    <div class="wp-pagenavi" role="navigation">
                      {% for page in info.pages %}
                      <a href="/list/{{page.page}}/"
                        class="{%if loop.first%}first{%endif%} {%if loop.last%}last{%endif%} {%if page.current%}current{%endif%}">
                        <span class="mdc-fab__icon">
                          {{page.text}}
                        </span>
                      </a>
                      {% endfor %}
                    </div>
                  </div>
                </div>
              </div>
              <div class="pages">

              </div>
            </div>
            <div
              class="mdc-layout-grid__cell mdc-layout-grid__cell--span-4 mdc-layout-grid__cell--span-12-tablet mdc-layout-grid__cell--span-12-phone">
              {% if info.slist.length > 0 %}
              <div class="listWidget">
                <div class="box">
                  <div class="title bread-crumbs">
                    <h5 class="widgetHeader">Hot Searches</h5>
                  </div>
                </div>
                <div>
                  <div class="appRow">
                    {% for item in info.slist %}
                    <a href="/search/{{item.ekey}}/" class="mdc-button">
                      <span class="mdc-button__label">{{item.key}}</span>
                    </a>
                    {% endfor %}
                  </div>
                </div>
              </div>
              {% endif %}
              <div class="listWidget">
                <div class="box">
                  <div class="title bread-crumbs">
                    <h5 class="widgetHeader">Social Medias</h5>
                  </div>
                </div>
                <div>
                  <div class="appRow">
                    <div class="table-row">
                      <div class="table-cell center">
                        <a href="https://www.facebook.com/profile.php?id=100071468057179" target="_blank">
                          <img style="vertical-align: baseline; height: 24px; width: 24px;" src="/img/facebook.png" class="clickable">
                        </a>
                      </div>
                      <div class="table-cell center">
                        <a href="https://twitter.com/Shaoyang2000" target="_blank">
                          <img style="vertical-align: baseline; height: 24px; width: 24px;" src="/img/twitter.png" class="clickable">
                        </a>
                      </div>
                      <div class="table-cell center">
                        <a href="https://t.me/idoras_official" target="_blank">
                          <img style="vertical-align: baseline; height: 24px; width: 24px;" src="/img/telegram.png" class="clickable">
                        </a>
                      </div>
                    </div>
                  </div>
                </div>
              </div>
              <div class="listWidget">
                <div class="box">
                  <div class="title bread-crumbs">
                    <h5 class="widgetHeader">Popular In Last 24 Hours</h5>
                  </div>
                </div>
                {% for item in info.hots[0].list %}
                <div>
                  <div class="appRow">
                    <div class="table-row">
                      <div style="width: 56px;" class="table-cell">
                        <img style="width:50px; height:50px;"
                          src="https://img.idoras.com/img/{{item.id}}.png">
                      </div>
                      <div class="table-cell">
                        <h5 title="{{item.name}} {{item.versionName}}" class="appRowTitle wrapText marginZero block-on-mobile">
                          <a class="fontBlack" href="/apk/{{item.id}}/">{{item.name}} {{item.versionName}}</a>
                        </h5>
                      </div>
                      <div style="width: 50px; text-align: right;" class="table-cell">
                        <span style="padding-right:12px;" class="colorLightBlack">{{item.down}}</span>
                      </div>
                      <div style="width: 22px;" class="table-cell">
                        <div class="iconsBox one-icon">
                          <div class="downloadIconPositioning">
                            <a class="downloadLink"
                              href="/apk/{{item.id}}/">
                              <i class="material-icons icon-blue" aria-hidden="true">download</i>
                            </a>
                          </div>
                        </div>
                      </div>
                    </div>
                  </div>
                </div>
                {% endfor %}
              </div>
              <div class="listWidget">
                <div class="box">
                  <div class="title bread-crumbs">
                    <h5 class="widgetHeader">Popular In Last 7 Days</h5>
                  </div>
                </div>
                {% for item in info.hots[1].list %}
                <div>
                  <div class="appRow">
                    <div class="table-row">
                      <div style="width: 56px;" class="table-cell">
                        <img style="width:50px; height:50px;"
                          src="https://img.idoras.com/img/{{item.id}}.png">
                      </div>
                      <div class="table-cell">
                        <h5 title="{{item.name}} {{item.versionName}}" class="appRowTitle wrapText marginZero block-on-mobile">
                          <a class="fontBlack" href="/apk/{{item.id}}/">{{item.name}} {{item.versionName}}</a>
                        </h5>
                      </div>
                      <div style="width: 50px; text-align: right;" class="table-cell">
                        <span style="padding-right:12px;" class="colorLightBlack">{{item.down}}</span>
                      </div>
                      <div style="width: 22px;" class="table-cell">
                        <div class="iconsBox one-icon">
                          <div class="downloadIconPositioning">
                            <a class="downloadLink"
                              href="/apk/{{item.id}}/">
                              <i class="material-icons icon-blue" aria-hidden="true">download</i>
                            </a>
                          </div>
                        </div>
                      </div>
                    </div>
                  </div>
                </div>
                {% endfor %}
              </div>
              <div class="listWidget">
                <div class="box">
                  <div class="title bread-crumbs">
                    <h5 class="widgetHeader">Popular In Last 30 Days</h5>
                  </div>
                </div>
                {% for item in info.hots[2].list %}
                <div>
                  <div class="appRow">
                    <div class="table-row">
                      <div style="width: 56px;" class="table-cell">
                        <img style="width:50px; height:50px;"
                          src="https://img.idoras.com/img/{{item.id}}.png">
                      </div>
                      <div class="table-cell">
                        <h5 title="{{item.name}} {{item.versionName}}" class="appRowTitle wrapText marginZero block-on-mobile">
                          <a class="fontBlack" href="/apk/{{item.id}}/">{{item.name}} {{item.versionName}}</a>
                        </h5>
                      </div>
                      <div style="width: 50px; text-align: right;" class="table-cell">
                        <span style="padding-right:12px;" class="colorLightBlack">{{item.down}}</span>
                      </div>
                      <div style="width: 22px;" class="table-cell">
                        <div class="iconsBox one-icon">
                          <div class="downloadIconPositioning">
                            <a class="downloadLink"
                              href="/apk/{{item.id}}/">
                              <i class="material-icons icon-blue" aria-hidden="true">download</i>
                            </a>
                          </div>
                        </div>
                      </div>
                    </div>
                  </div>
                </div>
                {% endfor %}
              </div>
            </div>
          </div>
        </div>
      </div>
      <div class="footer">
        <div class="b">
          <p><a href="/about/">About Us</a> | <a href="/contact-us/">Contact Us</a></p>
          <p><a href="/privacy-policy/" rel="nofollow">Privacy Policy</a></p>
          <p><a href="/report-content/">Report abuse</a></p>
          <p><a href="/dmca/" rel="nofollow">DMCA Disclaimer</a></p>
          <p>idoras.com © 2016-2021</p>
        </div>
      </div>
    </main>
  </div>
  <script src="/js/material-components-web.min.js"></script>
  <script src="/js/app.js?v=20210805"></script>
</script>
</body>

</html>

That’s the main code i wrote for the website, leave a comment if you want to know more about my code, at last remember the website address https://idoras.com i built a week ago, and share it with your firends if you find it useful, thanks.

字节跳动最常考的 64 道JS算法题

fengmao — Fri, 09 Apr 2021 13:08:22 +0000

缘起

现在大厂面试中，算法题几乎为必考项，且近几年频现 LeetCode 真题，此篇为拿到字节、腾讯、京东 Offer 的笔者本人在准备面试过程中亲自刷过以及遇到过高频算法题。文章内容会分模块整理，对于笔者在面试过程中遇到的真题，会给予着重【🔥】标出。

同时，可以毫不客气的说，如果你准备时间有限，又想追求算法题准备效率最大化，那么你只需要按照大纲把下面的题目刷完，并把代码烂熟于心，就几乎可以应对 90% 的面试算法考题了。

整理这篇内容的目的一个是笔者在之前准备面试时的一点积累，而它确实也帮助笔者在面试算法题中过关斩将，同时呢，也希望能够在金三银四给予拼搏的你，一点点帮助就好！💪

文末有福利：）😈

本篇内容包括如下模块：

高频算法题系列：链表
【🔥】【有真题】高频算法题系列：字符串
【🔥】【有真题】高频算法题系列：数组问题
高频算法题系列：二叉树
【🔥】高频算法题系列：排序算法
【🔥】高频算法题系列：二分查找
【🔥】高频算法题系列：动态规划
高频算法题系列：BFS
【🔥】高频算法题系列：栈
【🔥】高频算法题系列：DFS
【🔥】高频算法题系列：回溯算法

其中标🔥的部分代表非常高频的考题，其中不乏笔者遇到的原题。其中对于每一类，首先会列出包含的考题，然后针对每一道考题会给出难度、考察知识点、是否是面试真题，在每道题详细介绍时，还会给出每道题的 LeetCode 链接，帮助读者理解题意，以及能够进行实际的测验，还可以观看其他人的答案，更好的帮助准备。

高频算法题系列：链表

笔者遇到的高频链表题主要包含这几道：

通过链表的后续遍历判断回文链表问题【简单】
链表的反向输出【简单】
合并 K 个升序链表【困难】
K个一组翻转链表【困难】
环形链表【简单】
排序链表【中等】
相交链表【简单】

前序遍历判断回文链表

👉 【LeetCode 直通车】：234 回文链表（简单）

题解1

利用链表的后续遍历，使用函数调用栈作为后序遍历栈，来判断是否回文

→点击展开查看


/**
  *
  */
var isPalindrome = function(head) {
    let left = head;
    function traverse(right) {
        if (right == null) return true;
        let res = traverse(right.next);
        res = res && (right.val === left.val);
        left = left.next;
        return res;
    }
    return traverse(head);
};

复制代码

题解2

通过快、慢指针找链表中点，然后反转链表，比较两个链表两侧是否相等，来判断是否是回文链表

→点击展开查看


/**
  *
  */
var isPalindrome = function(head) {
    // 反转 slower 链表
    let right = reverse(findCenter(head));
    let left = head;
    // 开始比较
    while (right != null) {
        if (left.val !== right.val) {
            return false;
        }
        left = left.next;
        right = right.next;
    }
    return true;
}
function findCenter(head) {
    let slower = head, faster = head;
    while (faster && faster.next != null) {
        slower = slower.next;
        faster = faster.next.next;
    }
    // 如果 faster 不等于 null，说明是奇数个，slower 再移动一格
    if (faster != null) {
        slower = slower.next;
    }
    return slower;
}
function reverse(head) {
    let prev = null, cur = head, nxt = head;
    while (cur != null) {
        nxt = cur.next;
        cur.next = prev;
        prev = cur;
        cur = nxt;
    }
    return prev;
}

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反转链表

👉 【LeetCode 直通车】：206 反转链表（简单）

题解

→点击展开查看


/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function(head) {
    if (head == null || head.next == null) return head;
    let last = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return last;
};

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合并K个升序链表

👉 【LeetCode 直通车】：23 合并K个升序链表（困难）

题解

→点击展开查看


/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode[]} lists
 * @return {ListNode}
 */
var mergeKLists = function(lists) {
    if (lists.length === 0) return null;
    return mergeArr(lists);
};
function mergeArr(lists) {
    if (lists.length <= 1) return lists[0];
    let index = Math.floor(lists.length / 2);
    const left = mergeArr(lists.slice(0, index))
    const right = mergeArr(lists.slice(index));
    return merge(left, right);
}
function merge(l1, l2) {
    if (l1 == null && l2 == null) return null;
    if (l1 != null && l2 == null) return l1;
    if (l1 == null && l2 != null) return l2;
    let newHead = null, head = null;
    while (l1 != null && l2 != null) {
        if (l1.val < l2.val) {
            if (!head) {
                newHead = l1;
                head = l1;
            } else {
                newHead.next = l1;
                newHead = newHead.next;
            }
            l1 = l1.next;
        } else {
            if (!head) {
                newHead = l2;
                head = l2;
            } else {
                newHead.next = l2;
                newHead = newHead.next;
            }
            l2 = l2.next;
        }
    }
    newHead.next = l1 ? l1 : l2;
    return head;
}

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K 个一组翻转链表

👉 【LeetCode 直通车】：25 K 个一组翻转链表（困难）

题解

→点击展开查看


/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
var reverseKGroup = function(head, k) {
    let a = head, b = head;
    for (let i = 0; i < k; i++) {
        if (b == null) return head;
        b = b.next;
    }
    const newHead = reverse(a, b);
    a.next = reverseKGroup(b, k);
    return newHead;
};
function reverse(a, b) {
    let prev = null, cur = a, nxt = a;
    while (cur != b) {
        nxt = cur.next;
        cur.next = prev;
        prev = cur;
        cur = nxt;
    }
    return prev;
}

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环形链表

👉 【LeetCode 直通车】：141 环形链表（简单）

题解

→点击展开查看


/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function(head) {
    if (head == null || head.next == null) return false;
    let slower = head, faster = head;
    while (faster != null && faster.next != null) {
        slower = slower.next;
        faster = faster.next.next;
        if (slower === faster) return true;
    }
    return false;
};

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排序链表

👉 【LeetCode 直通车】：148 排序链表（中等）

题解

→点击展开查看


/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var sortList = function(head) {
    if (head == null) return null;
    let newHead = head;
    return mergeSort(head);
};
function mergeSort(head) {
    if (head.next != null) {
        let slower = getCenter(head);
        let nxt = slower.next;
        slower.next = null;
        console.log(head, slower, nxt);
        const left = mergeSort(head);
        const right = mergeSort(nxt);
        head = merge(left, right);
    }
    return head;
}
function merge(left, right) {
    let newHead = null, head = null;
    while (left != null && right != null) {
        if (left.val < right.val) {
            if (!head) {
                newHead = left;
                head = left;
            } else {
                newHead.next = left;
                newHead = newHead.next;
            }
            left = left.next;
        } else {
            if (!head) {
                newHead = right;
                head = right;
            } else {
                newHead.next = right;
                newHead = newHead.next;
            }
            right = right.next;
        }
    }
    newHead.next = left ? left : right;
    return head;
}
function getCenter(head) {
    let slower = head, faster = head.next;
    while (faster != null && faster.next != null) {
        slower = slower.next;
        faster = faster.next.next;
    }
    return slower;
}

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相交链表

👉 【LeetCode 直通车】：160 相交链表（简单）

题解

→点击展开查看


/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function(headA, headB) {
    let lastHeadA = null;
    let lastHeadB = null;
    let originHeadA = headA;
    let originHeadB = headB;
    if (!headA || !headB) {
        return null;
    }
    while (true) {
        if (headB == headA) {
            return headB;
        }
        if (headA && headA.next == null) {
            lastHeadA = headA;
            headA = originHeadB;
        } else {
            headA = headA.next;
        }
        if (headB && headB.next == null) {
            lastHeadB = headB
            headB = originHeadA;
        } else {
            headB = headB.next;
        }
        if (lastHeadA && lastHeadB && lastHeadA != lastHeadB) {
            return null;
        }
    }
    return null;
};

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【🔥】高频算法题系列：字符串

主要有以下几类高频考题：

最长回文子串【中等】【双指针】【面试真题】
最长公共前缀【简单】【双指针】
无重复字符的最长子串【中等】【双指针】
最小覆盖子串【困难】【滑动窗口】【面试真题】

【面试真题】最长回文子串【双指针】

👉 【LeetCode 直通车】：5 最长回文子串（中等）

题解

→点击展开查看


/**
 * @param {string} s
 * @return {string}
 */
var longestPalindrome = function(s) {
    if (s.length === 1) return s;
    let maxRes = 0, maxStr = '';
    for (let i = 0; i < s.length; i++) {
        let str1 = palindrome(s, i, i);
        let str2 = palindrome(s, i, i + 1);   
        if (str1.length > maxRes) {
            maxStr = str1;
            maxRes = str1.length;
        }
        if (str2.length > maxRes) {
            maxStr = str2;
            maxRes = str2.length;
        }
    }
    return maxStr;
};
function palindrome(s, l, r) {
    while (l >= 0 && r < s.length && s[l] === s[r]) {
        l--;
        r++;
    }
    return s.slice(l + 1, r);
}

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最长公共前缀【双指针】

👉 【LeetCode 直通车】：14 最长公共前缀（简单）

题解

→点击展开查看


/**
 * @param {string[]} strs
 * @return {string}
 */
var longestCommonPrefix = function(strs) {
    if (strs.length === 0) return "";
    let first = strs[0];
    if (first === "") return "";
    let minLen = Number.MAX_SAFE_INTEGER;
    for (let i = 1; i < strs.length; i++) {
        const len = twoStrLongestCommonPrefix(first, strs[i]);
        minLen = Math.min(len, minLen);
    }
    return first.slice(0, minLen);
};
function twoStrLongestCommonPrefix (s, t) {
    let i = 0, j = 0;
    let cnt = 0;
    while (i < s.length && j < t.length) {
        console.log(s[i], t[j], cnt)
        if (s[i] === t[j])  {
            cnt++;
        } else {
            return cnt;
        }
        i++;
        j++;
    }
    return cnt;
}

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无重复字符的最长子串【双指针】

👉 【LeetCode 直通车】：3 无重复字符的最长子串（中等）

题解

→点击展开查看


/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function(s) {
  let window = {};
  let left = 0, right = 0;
  let maxLen = 0, maxStr = '';
  while (right < s.length) {
    let c = s[right];
    right++;
    if (window[c]) window[c]++;
    else window[c] = 1
    while (window[c] > 1) {
      let d = s[left];
      left++;
      window[d]--;
    }
    if (maxLen < right - left) {
      maxLen = right - left;
    }
  }
  return maxLen;
};

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【面试真题】最小覆盖子串【滑动窗口】

👉 【LeetCode 直通车】：76 最小覆盖子串（困难）

题解

→点击展开查看


/**
 * @param {string} s
 * @param {string} t
 * @return {string}
 */
var minWindow = function(s, t) {
    let need = {}, window = {};
    for (let c of t) {
        if (!need[c]) need[c] = 1;
        else need[c]++;
    }
    let left = 0, right = 0;
    let valid = 0, len = Object.keys(need).length;
    let minLen = s.length + 1, minStr = '';
    while (right < s.length) {
        const d = s[right];
        right++;
        if (!window[d]) window[d] = 1;
        else window[d]++;
        if (need[d] && need[d] === window[d]) {
            valid++;
        }
        console.log('left - right', left, right);
        while (valid === len) {
            if (right - left < minLen) {
                minLen = right - left;
                minStr = s.slice(left, right);
            }
            console.lo
            let c = s[left];
            left++;
            window[c]--;
            if (need[c] && window[c] < need[c]) {
                valid--;
            }
        }
    }
    return minStr;
};

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【🔥】高频算法题系列：数组问题

主要有几类高频考题：

俄罗斯套娃信封问题【困难】【排序+最长上升子序列】【面试真题】
最长连续递增序列【简单】【双指针】
最长连续序列【困难】【哈希表】
盛最多水的容器【困难】【面试真题】
寻找两个正序数组的中位数【困难】【双指针】
删除有序数组中的重复项【简单】【快慢指针】
和为K的子数组【中等】【哈希表】
nSum 问题【系列】【简单】【哈希表】
接雨水【困难】【暴力+备忘录优化】【面试真题】
跳跃游戏【系列】【中等】【贪心算法】

【面试真题】俄罗斯套娃信封问题【排序+最长上升子序列】

👉 【LeetCode 直通车】：354 俄罗斯套娃信封问题（困难）

题解

→点击展开查看


/**
 * @param {number[][]} envelopes
 * @return {number}
 */
var maxEnvelopes = function(envelopes) {
  if (envelopes.length === 1) return 1;
    envelopes.sort((a, b) => {
        if (a[0] !== b[0]) return a[0] - b[0];
        else return b[1] - a[1];
    });
    let LISArr = [];
    for (let [key, value] of envelopes) {
      LISArr.push(value);
    }
    console.log( LISArr);
    return LIS(LISArr);
};
function LIS(arr) {
  let dp = [];
  let maxAns = 0;
  for (let i = 0; i < arr.length; i++) {
    dp[i] = 1;
  }
  for (let i = 1; i < arr.length; i++) {
    for (let j = i; j >= 0; j--) {
      if (arr[i] > arr[j]) {
        dp[i] = Math.max(dp[i], dp[j] + 1)
      }
      maxAns = Math.max(maxAns, dp[i]);
    }
  }
  return maxAns;
}

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最长连续递增序列【快慢指针】

👉 【LeetCode 直通车】：674 最长连续递增序列（简单）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {number}
 */
var findLengthOfLCIS = function(nums) {
    if (nums.length === 0) return 0;
    const n = nums.length;
    let left = 0, right = 1;
    let globalMaxLen = 1, maxLen = 1;
    while (right < n) {
        if (nums[right] > nums[left]) maxLen++;
        else {
            maxLen = 1;
        }
        left++;
        right++;
        globalMaxLen = Math.max(globalMaxLen, maxLen);
    }
    return globalMaxLen;
};

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最长连续序列【哈希表】

👉 【LeetCode 直通车】：128 最长连续序列（困难）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {number}
 */
var longestConsecutive = function(nums) {
    if (nums.length === 0) return 0;
    const set = new Set(nums);
    const n = nums.length;
    let globalLongest = 1;
    for (let i = 0; i < n; i++) {
        if (!set.has(nums[i] - 1)) {
            let longest = 1;
            let currentNum = nums[i];
            while (set.has(currentNum + 1)) {
                currentNum += 1;
                longest++;
            }
            globalLongest = Math.max(globalLongest, longest);
        }
    }
    return globalLongest;
};

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【面试真题】盛最多水的容器【哈希表】

👉 【LeetCode 直通车】：11 盛最多水的容器（中等）

题解

→点击展开查看


/**
 * @param {number[]} height
 * @return {number}
 */
var maxArea = function(height) {
    let n = height.length;
    let left = 0, right = n - 1;
    let maxOpacity = 0;
    while (left < right) {
        let res = Math.min(height[left], height[right]) * (right - left);
        maxOpacity = Math.max(maxOpacity, res);
        if (height[left] < height[right]) left++
        else right--;
    }
    return maxOpacity;
};

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寻找两个正序数组的中位数【双指针】

👉 【LeetCode 直通车】：4 寻找两个正序数组的中位数（困难）

题解

→点击展开查看


/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function(nums1, nums2) {
    let m = nums1.length, n = nums2.length;
    let i = 0, j = 0;
    let newArr = [];
    while (i < m && j < n) {
        if (nums1[i] < nums2[j]) {
            newArr.push(nums1[i++]);
        } else {
            newArr.push(nums2[j++]);
        }
    }
    newArr = newArr.concat(i < m ? nums1.slice(i) : nums2.slice(j));
    const len = newArr.length;
    console.log(newArr)
    if (len % 2 === 0) {
        return (newArr[len / 2] + newArr[len / 2 - 1]) / 2;
    } else {
        return newArr[Math.floor(len / 2)];
    }
};

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删除有序数组中的重复项【快慢指针】

👉 【LeetCode 直通车】：26 删除有序数组中的重复项（简单）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
  if (nums.length <= 1) return nums.length;
  let lo = 0, hi = 0;
  while (hi < nums.length) {
    while (nums[lo] === nums[hi] && hi < nums.length) hi++;
    if (nums[lo] !== nums[hi] && hi < nums.length) {
      lo++;
      nums[lo] = nums[hi];
    }
    hi++;
  }
  return lo + 1;
};

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和为K的子数组【哈希表】

👉 【LeetCode 直通车】：560 和为K的子数组（中等）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var subarraySum = function(nums, k) {
    let cnt = 0;
    let sum0_i = 0, sum0_j = 0;
    let map = new Map();
    map.set(0, 1);
    for (let i = 0; i <= nums.length; i++) {
        sum0_i += nums[i];
        sum0_j = sum0_i - k;
        console.log('map', sum0_j, map.get(sum0_j))
        if (map.has(sum0_j)) {
            cnt += map.get(sum0_j);
        }
        let sumCnt = map.get(sum0_i) || 0;
        map.set(sum0_i, sumCnt + 1);
    }
    return cnt;
};

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nSum问题【哈希表】【系列】

受限于篇幅，这里只给出第一道题的代码模板，也是一面常考真题。

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
  let map2 = new Map();
  for (let i = 0; i < nums.length; i++) {
    map2.set(nums[i], i);
  }
  for (let i = 0; i < nums.length; i++) {
    if (map2.has(target - nums[i]) && map2.get(target - nums[i]) !== i) return [i, map2.get(target - nums[i])]
  }
};

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【面试真题】接雨水【暴力+备忘录优化】

👉 【LeetCode 直通车】：42 接雨水（困难）

题解

→点击展开查看


/**
 * @param {number[]} height
 * @return {number}
 */
var trap = function(height) {
    let l_max = [], r_max = [];
    let len = height.length;
    let maxCapacity = 0;
    for (let i = 0; i < len; i++) {
        l_max[i] = height[i];
        r_max[i] = height[i];
    }
    for (let i = 1; i < len; i++) {
        l_max[i] = Math.max(l_max[i - 1], height[i]);
    }
    for (let j = len - 2; j >= 0; j--) {
        r_max[j] = Math.max(r_max[j + 1], height[j]);
    }
    for (let i = 0; i < len; i++) {
        maxCapacity += Math.min(l_max[i], r_max[i]) - height[i];
    }
    return maxCapacity;
};

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跳跃游戏【贪心算法】【系列】

受限于篇幅，这里只给出第一道题的代码模板，也是一面常考真题。

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {boolean}
 */
var canJump = function(nums) {
    let faster = 0;
    for (let i = 0; i < nums.length - 1; i++) {
        faster = Math.max(faster, i + nums[i]);
        if (faster <= i) return false;
    }
    return faster >= nums.length - 1;
};

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高频算法题系列：二叉树

主要有以下几类高频考题：

二叉树的最近公共祖先【简单】【二叉树】
二叉搜索树中的搜索【简单】【二叉树】
删除二叉搜索树中的节点【中等】【二叉树】
完全二叉树的节点个数【中等】【二叉树】
二叉树的锯齿形层序遍历【中等】【二叉树】

二叉树的最近公共祖先【二叉树】

👉 【LeetCode 直通车】：236 二叉树的最近公共祖先（简单）

题解

→点击展开查看


/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
let visited;let parent;
var lowestCommonAncestor = function(root, p, q) {
    visited = new Set();
    parent = new Map();
    dfs(root);
    while (p != null) {
        visited.add(p.val);
        p = parent.get(p.val);
    }
    while (q != null) {
        if (visited.has(q.val)) {
            return q;
        }
        q = parent.get(q.val);
    }
    return null;
};
function dfs(root) {
    if (root.left != null) {
        parent.set(root.left.val, root);
        dfs(root.left);
    }
    if (root.right != null) {
        parent.set(root.right.val, root);
        dfs(root.right);
    }
}

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二叉搜索树中的搜索【二叉树】

👉 【LeetCode 直通车】：700 二叉搜索树中的搜索（简单）

题解

→点击展开查看


/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} val
 * @return {TreeNode}
 */
var searchBST = function(root, val) {
    if (root == null) return null;
    if (root.val === val) return root;
    if (root.val > val) {
        return searchBST(root.left, val);
    } else if (root.val < val) {
        return searchBST(root.right, val);
    }
};

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删除二叉搜索树中的节点【二叉树】

👉 【LeetCode 直通车】：450 删除二叉搜索树中的节点（中等）

题解

→点击展开查看


/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} key
 * @return {TreeNode}
 */
var deleteNode = function(root, key) {
    if (root == null) return null;
    if (root.val === key) {
        if (root.left == null && root.right == null) return null;
        if (root.left == null) return root.right;
        if (root.right == null) return root.left;
        if (root.left != null && root.right != null)  {
            let target = getMinTreeMaxNode(root.left);
            root.val = target.val;
            root.left = deleteNode(root.left, target.val);
        }
    }
    if (root.val < key) {
        root.right = deleteNode(root.right, key);
    } else if (root.val > key) {
        root.left = deleteNode(root.left, key);
    }
    return root;
};
function getMinTreeMaxNode(root) {
    if (root.right == null) return root;
    return getMinTreeMaxNode(root.right);
}

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完全二叉树的节点个数【二叉树】

👉 【LeetCode 直通车】：222 完全二叉树的节点个数（中等）

题解

→点击展开查看


/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var countNodes = function(root) {
    if (root == null) return 0;
    let l = root, r = root;
    let lh = 0, rh = 0;
    while (l != null) {
      l = l.left;
      lh++;
    }
    while (r != null) {
      r = r.right;
      rh++;
    }
    if (lh === rh) {
      return Math.pow(2, lh) - 1;
    }
    return 1 + countNodes(root.left) + countNodes(root.right);
};

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二叉树的锯齿形层序遍历【二叉树】

👉 【LeetCode 直通车】：103 二叉树的锯齿形层序遍历（中等）

题解

→点击展开查看


/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
let res;
var zigzagLevelOrder = function(root) {
    if (root == null) return [];
    res = [];
    BFS(root, true);
    return res;
};
function BFS(root, inOrder) {
    let arr = [];
    let resItem = [];
    let node;
    let stack1 = new Stack();
    let stack2 = new Stack();
    // 判断交换时机
    let flag;
    stack1.push(root);
    res.push([root.val]);
    inOrder = !inOrder;
    while (!stack1.isEmpty() || !stack2.isEmpty()) {
        if (stack1.isEmpty()) {
            flag = 'stack1';
        } else if (stack2.isEmpty()) {
            flag = 'stack2';
        }
        // 决定取那个栈里面的元素
        if (flag === 'stack2' && !stack1.isEmpty()) node = stack1.pop();
        else if (flag === 'stack1' && !stack2.isEmpty()) node = stack2.pop();
        if (inOrder) {
            if (node.left) {
                if (flag === 'stack1') {
                    stack1.push(node.left);
                } else {
                    stack2.push(node.left);
                }
                resItem.push(node.left.val);
            }
            if (node.right) {
                if (flag === 'stack1') {
                    stack1.push(node.right);
                } else {
                    stack2.push(node.right);
                }
                resItem.push(node.right.val);
            }
        } else {
            if (node.right) {
                if (flag === 'stack1') {
                    stack1.push(node.right);
                } else {
                    stack2.push(node.right);
                }
                resItem.push(node.right.val);
            }
            if (node.left) {
                if (flag === 'stack1') {
                    stack1.push(node.left);
                } else {
                    stack2.push(node.left);
                }
                resItem.push(node.left.val);
            }
        }
        // 判断下次翻转的时机
        if ((flag === 'stack2' && stack1.isEmpty()) || (flag === 'stack1' && stack2.isEmpty())) {
            inOrder = !inOrder;
            // 需要翻转了，就加一轮值
            if (resItem.length > 0) {
                res.push(resItem);
            }   
            resItem = [];
        }
    }
}
class Stack {
    constructor() {
        this.count = 0;
        this.items = [];
    }
    push(element) {
        this.items[this.count] = element;
        this.count++;
    }
    pop() {
        if (this.isEmpty()) return undefined;
        const element = this.items[this.count - 1];
        delete this.items[this.count - 1];
        this.count--;
        return element;
    }
    size() {
        return this.count;
    }
    isEmpty() {
        return this.size() === 0;
    }
}

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【🔥】高频算法题系列：排序算法

主要有以下几类高频考题：

用最少数量的箭引爆气球【中等】【排序】
合并区间【中等】【排序算法+区间问题】【面试真题】

用最少数量的箭引爆气球【排序算法】

👉 【LeetCode 直通车】：452 用最少数量的箭引爆气球（中等）

题解

→点击展开查看


/**
 * @param {number[][]} points
 * @return {number}
 */
var findMinArrowShots = function(points) {
    if (points.length === 0) return 0;
    points.sort((a, b) => a[1] - b[1]);
    let cnt = 1;
    let resArr = [points[0]];
    let curr, last;
    for (let i = 1; i < points.length; i++) {
        curr = points[i];
        last = resArr[resArr.length - 1];
        if (curr[0] > last[1]) {
            resArr.push(curr);
            cnt++;
        }
    }
    return cnt;
};

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合并区间【排序算法+区间问题】

👉 【LeetCode 直通车】：56 合并区间（中等）

题解

→点击展开查看


/**
 * @param {number[][]} intervals
 * @return {number[][]}
 */
var merge = function(intervals) {
    if (intervals.length === 0) return [];
    intervals.sort((a, b) => a[0] - b[0]);
    let mergeArr = [intervals[0]];
    let last, curr;
    for (let j = 1; j < intervals.length; j++) {
        last = mergeArr[mergeArr.length - 1];
        curr = intervals[j];
        if (last[1] >= curr[0]) {
            last[1] = Math.max(curr[1], last[1]);
        } else {
            mergeArr.push(curr);
        }
    }
    return mergeArr;
};

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高频算法题系列：二分查找

主要有以下几类高频考题：

寻找两个正序数组的中位数【困难】【二分查找】
判断子序列【简单】【二分查找】
在排序数组中查找元素的第一个和最后一个位置【中等】【二分查找】

寻找两个正序数组的中位数【二分查找】

👉 【LeetCode 直通车】：4 寻找两个正序数组的中位数（困难）

题解

→点击展开查看


/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function(nums1, nums2) {
    let m = nums1.length, n = nums2.length;
    let i = 0, j = 0;
    let newArr = [];
    while (i < m && j < n) {
        if (nums1[i] < nums2[j]) {
            newArr.push(nums1[i++]);
        } else {
            newArr.push(nums2[j++]);
        }
    }
    newArr = newArr.concat(i < m ? nums1.slice(i) : nums2.slice(j));
    const len = newArr.length;
    console.log(newArr)
    if (len % 2 === 0) {
        return (newArr[len / 2] + newArr[len / 2 - 1]) / 2;
    } else {
        return newArr[Math.floor(len / 2)];
    }
};

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判断子序列【二分查找】

👉 【LeetCode 直通车】：392 判断子序列（简单）

题解

→点击展开查看


/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isSubsequence = function(s, t) {
    let hash = {};
    for (let i = 0; i < t.length; i++) {
        if (!hash[t[i]]) hash[t[i]] = [];
        hash[t[i]].push(i);
    }
    let lastMaxIndex = 0;
    for (let i = 0; i < s.length; i++) {
        if (hash[s[i]]) {
            const index = binarySearch(hash[s[i]], lastMaxIndex);
            console.log('index', index, hash[s[i]]);
            if (index === -1) return false;
            lastMaxIndex = hash[s[i]][index] + 1;
        } else return false;
    }
    return true;
};
function binarySearch(array, targetIndex) {
    let left = 0, right = array.length;
    while (left < right) {
        let mid = left + Math.floor((right - left) / 2);
        if (array[mid] >= targetIndex) {
            right = mid;
        } else if (array[mid] < targetIndex) {
            left = mid + 1;
        }
    }
    if (left >= array.length || array[left] < targetIndex) return -1;
    return left;
}

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💁 在排序数组中查找元素的第一个和最后一个位置【二分搜索】

👉 【LeetCode 直通车】：34 在排序数组中查找元素的第一个和最后一个位置（中等）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var searchRange = function(nums, target) {
    const left = leftBound(nums, target);
    const right = rightBound(nums, target);
    return [left, right];
};
function leftBound(nums, target) {
    let left = 0;
    let right = nums.length - 1;
    while (left <= right) {
        let mid = Math.floor(left + (right - left) / 2);
        if (nums[mid] === target) {
            right = mid - 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1;
        }
    }
    if (left >= nums.length || nums[left] !== target) {
        return -1;
    }
    return left;
}
function rightBound(nums, target) {
    let left = 0;
    let right = nums.length - 1;
    while (left <= right) {
        let mid = Math.floor(left + (right - left) / 2);
        if (nums[mid] === target) {
            left = mid + 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1;
        }
    }
    if (right < 0 || nums[right] !== target) {
        return -1;
    }
    return right;
}

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【🔥】高频算法题系列：动态规划

主要有以下几类高频考题：

最长递增子序列【中等】【动态规划】
零钱兑换【中等】【动态规划】【面试真题】
最长公共子序列【中等】【动态规划】【面试真题】
编辑距离【困难】【动态规划】
最长回文子序列【中等】【动态规划】【面试真题】
最大子序和【简单】【动态规划】【面试真题】
买卖股票的最佳时机系列【系列】【动态规划】【面试真题】

最长递增子序列【动态规划】

👉 【LeetCode 直通车】：300 最长递增子序列（中等）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {number}
 */
var lengthOfLIS = function(nums) {
    let maxLen = 0, n = nums.length;
    let dp = [];
    for (let i = 0; i < n; i++) {
        dp[i] = 1;
    }
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < i; j++) {
            if (nums[i] > nums[j]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
        maxLen = Math.max(maxLen, dp[i]);
    }
    return maxLen;
};

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【面试真题】零钱兑换【动态规划】

👉 【LeetCode 直通车】：322 零钱兑换（中等）

题解

→点击展开查看


/**
 * @param {number[]} coins
 * @param {number} amount
 * @return {number}
 */
var coinChange = function(coins, amount) {
  if (amount === 0) return 0;
  let dp = [];
  for (let i = 0; i <= amount; i++) {
    dp[i] = amount + 1;
  }
  dp[0] = 0;
  for (let i = 0; i <= amount; i++) {
    for (let j = 0; j < coins.length; j++) {
      if (i >= coins[j]) {
        dp[i] = Math.min(dp[i - coins[j]] + 1, dp[i])
      }
    }
  }
  return dp[amount] === amount + 1 ? -1 : dp[amount];
};

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【面试真题】最长公共子序列【动态规划】

👉 【LeetCode 直通车】：1143 最长公共子序列（中等）

题解

→点击展开查看


/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function(text1, text2) {
    let n1 = text1.length, n2 = text2.length;
    let dp = [];
    for (let i = -1; i < n1; i++) {
        dp[i] = [];
        for (let j = -1; j < n2;j++) {
            dp[i][j] = 0;
        }
    }
    for (let i = 0; i < n1; i++) {
        for (let j = 0; j < n2; j++) {
            if (text1[i] === text2[j]) {
                dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 1] + 1);
            } else {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1])
            }
        }
    }
    return dp[n1 - 1][n2 - 1];
};

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编辑距离【动态规划】

👉 【LeetCode 直通车】：72 编辑距离（困难）

题解

→点击展开查看


/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
var minDistance = function(word1, word2) {
  let len1 = word1.length, len2 = word2.length;
  let dp = [];
  for (let i = 0; i <= len1; i++) {
    dp[i] = [];
    for (let j = 0; j <= len2; j++) {
      dp[i][j] = 0;
      if (i === 0) {
        dp[i][j] = j;
      }
      if (j === 0) {
        dp[i][j] = i;
      }
    }
  }
  for (let i = 1; i <= len1; i++) {
    for (let j = 1; j <= len2; j++) {
      if (word1[i - 1] === word2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1];
      } else {
        dp[i][j] = Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1] + 1);
      }
    }
  }
  return dp[len1][len2];
};

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【面试真题】最长回文子序列【动态规划】

👉 【LeetCode 直通车】：516 最长回文子序列（中等）

题解

→点击展开查看


/**
 * @param {string} s
 * @return {number}
 */
var longestPalindromeSubseq = function(s) {
    let dp = [];
    for (let i = 0; i < s.length; i++) {
        dp[i] = [];
        for (let j = 0; j < s.length; j++) {
            dp[i][j] = 0;
        }
        dp[i][i] = 1;
    }
    for (let i = s.length - 1; i >= 0; i--) {
        for (let j = i + 1; j < s.length; j++) {
            if (s[i] === s[j]) {
                dp[i][j] = dp[i + 1][j - 1] + 2;
            } else {
                dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
            }
        }
    }
    return dp[0][s.length - 1];
};

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【面试真题】💁 最大子序和【动态规划】

👉 【LeetCode 直通车】：53 最大子序和（简单）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {
    let maxSum = -Infinity;
    let dp = [], n = nums.length;
    for (let i = -1; i < n; i++) {
        dp[i] = 0;
    }
    for (let i = 0; i < n; i++) {
        dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]);
        maxSum = Math.max(maxSum, dp[i]);
    }
    return maxSum;
};

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【面试真题】💁 买卖股票的最佳时机【动态规划】

受限于篇幅，这里只给出第一道题的代码模板，也是一面常考真题，笔者在面试字节跳动时就遇到过。

题解

→点击展开查看


/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
  let dp = [];
  for (let i = -1; i < prices.length; i++) {
    dp[i] = []
    for (let j = 0; j <= 1; j++) {
      dp[i][j] = [];
      dp[i][j][0] = 0;
      dp[i][j][1] = 0;
      if (i === -1) {
        dp[i][j][1] = -Infinity;
      }
      if (j === 0) {
        dp[i][j][1] = -Infinity;
      }
      if (j === -1) {
        dp[i][j][1] = -Infinity;
      }
    }
  }
  for (let i = 0; i < prices.length; i++) {
    for (let j = 1; j <= 1; j++) {
      dp[i][j][0] = Math.max(dp[i - 1][j][0], dp[i - 1][j][1] + prices[i]);
      dp[i][j][1] = Math.max(dp[i - 1][j][1], dp[i - 1][j - 1][0] - prices[i]);
    }
  }
  return dp[prices.length - 1][1][0];
};

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高频算法题系列：BFS

主要有以下几类高频考题：

打开转盘锁【中等】【BFS】
二叉树的最小深度【简单】【BFS】

打开转盘锁【BFS】

👉 【LeetCode 直通车】：752 打开转盘锁（中等）

题解

→点击展开查看


/**
 * @param {string[]} deadends
 * @param {string} target
 * @return {number}
 */
var openLock = function(deadends, target) {
  let queue = new Queue();
  let visited = new Set();
  let step = 0;
  queue.push('0000');
  visited.add('0000');
  while (!queue.isEmpty()) {
    let size = queue.size();
    for (let i = 0; i < size; i++) {
      let str = queue.pop();
      if (deadends.includes(str)) continue;
      if (target === str) {
        return step;
      }
      for (let j = 0; j < 4; j++) {
        let plusStr = plusOne(str, j);
        let minusStr = minusOne(str, j);
        if (!visited.has(plusStr)) {
          queue.push(plusStr);
          visited.add(plusStr)
        }
        if (!visited.has(minusStr)) {
          queue.push(minusStr);
          visited.add(minusStr)
        }
      }
    }
    step++;
  }
  return -1;
};
function plusOne(str, index) {
  let strArr = str.split('');
  if (strArr[index] === '9') {
    strArr[index] = '0'
  } else {
    strArr[index] = (Number(strArr[index]) + 1).toString()
  }
  return strArr.join('');
}
function minusOne(str, index) {
  let strArr = str.split('');
  if (strArr[index] === '0') {
    strArr[index] = '9'
  } else {
    strArr[index] = (Number(strArr[index]) - 1).toString()
  }
  return strArr.join('');
}
class Queue {
  constructor() {
    this.items = [];
    this.count = 0;
    this.lowerCount = 0;
  }
  push(elem) {
    this.items[this.count++] = elem;
  }
  pop() {
    if (this.isEmpty()) {
      return;
    }
    const elem = this.items[this.lowerCount];
    delete this.items[this.lowerCount];
    this.lowerCount++;
    return elem;
  }
  isEmpty() {
    if (this.size() === 0) return true;
    return false;
  }
  size() {
    return this.count - this.lowerCount;
  }
}

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二叉树的最小深度【BFS】

👉 【LeetCode 直通车】：111 二叉树的最小深度（简单）

题解

→点击展开查看


/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function(root) {
  if (root == null) return 0;
  let depth = 1;
  let queue = new Queue();
  queue.push(root);
  while (!queue.isEmpty()) {
    let size = queue.size();
    for (let i = 0; i < size; i++) {
      const node = queue.pop();
      if (node.left == null && node.right == null) return depth;
      if (node.left) {
        queue.push(node.left);
      }
      if (node.right) {
        queue.push(node.right);
      }
    }
    depth++;
  }
  return depth;
};
class Queue {
  constructor() {
    this.items = [];
    this.count = 0;
    this.lowerCount = 0;
  }
  push(elem) {
    this.items[this.count++] = elem;
  }
  pop() {
    if (this.isEmpty()) {
      return;
    }
    const elem = this.items[this.lowerCount];
    delete this.items[this.lowerCount];
    this.lowerCount++;
    return elem;
  }
  isEmpty() {
    if (this.size() === 0) return true;
    return false;
  }
  size() {
    return this.count - this.lowerCount;
  }
}

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【🔥】高频算法题系列：栈

主要有以下几类高频考题：

最小栈【简单】【栈】
有效的括号【中等】【栈】【面试真题】
简化路径【中等】【栈】
下一个更大元素【系列】【栈】

最小栈【栈】

👉 【LeetCode 直通车】：155 最小栈（简单）

题解

→点击展开查看


/**
 * initialize your data structure here.
 */
var MinStack = function() {
    this.stack = [];
    this.minArr = [];
    this.count = 0;
    this.min = Number.MAX_SAFE_INTEGER;
};
/** 
 * @param {number} x
 * @return {void}
 */
MinStack.prototype.push = function(x) {
    this.min = Math.min(this.min, x);
    this.minArr[this.count] = this.min;
    this.stack[this.count] = x;
    this.count++;
};
/**
 * @return {void}
 */
MinStack.prototype.pop = function() {
    const element = this.stack[this.count - 1];
    if (this.count - 2 >= 0) this.min = this.minArr[this.count - 2];
    else  this.min = Number.MAX_SAFE_INTEGER;
    delete this.stack[this.count - 1];
    delete this.minArr[this.count - 1];
    this.count--;
    return element;
};
/**
 * @return {number}
 */
MinStack.prototype.top = function() {
    if (this.count >= 1) {
        return this.stack[this.count - 1];
    }
    return null;
};
/**
 * @return {number}
 */
MinStack.prototype.getMin = function() {
    const element = this.minArr[this.count - 1];
    return element;
};
/**
 * Your MinStack object will be instantiated and called as such:
 * var obj = new MinStack()
 * obj.push(x)
 * obj.pop()
 * var param_3 = obj.top()
 * var param_4 = obj.getMin()
 */

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【系列】下一个更大元素【栈】

受限于篇幅，这里只给出第一道题的代码模板

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {number[]}
 */
var nextGreaterElements = function(nums) {
    let ans = [];
    let stack = new Stack();
    const n = nums.length;
    for (let i = 2 * n - 1; i >= 0; i--) {
        while (!stack.isEmpty() && stack.top() <= nums[i % n]) {
            stack.pop();
        }
        ans[i % n] = stack.isEmpty() ? -1 : stack.top();
        stack.push(nums[i % n]);
    }
    return ans;
};
class Stack {
    constructor() {
        this.count = 0;
        this.items = [];
    }
    top() {
        if (this.isEmpty()) return undefined;
        return this.items[this.count - 1];
    }
    push(element) {
        this.items[this.count] = element;
        this.count++;
    }
    pop() {
        if (this.isEmpty()) return undefined;
        const element = this.items[this.count - 1];
        delete this.items[this.count - 1];
        this.count--;
        return element;
    }
    isEmpty() {
        return this.size() === 0;
    }
    size() {
        return this.count;
    }
}

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【面试真题】有效的括号【栈】

👉 【LeetCode 直通车】：20 有效的括号（中等）

题解

→点击展开查看


/**
 * @param {string} s
 * @return {boolean}
 */
var isValid = function(s) {
    if (s.length === 0) {
        return true;
    }
    if (s.length % 2 !== 0) {
        return false;
    }
    let map = {
        ')': '(',
        ']': '[',
        '}': '{',
    };
    let left = ['(', '[', '{'];
    let right = [')', ']', '}'];
    let stack = new Stack();
    for (let i = 0; i < s.length; i++) {
        if (!right.includes(s[i])) {
            stack.push(s[i]);
        } else {
            const matchStr = map[s[i]];
            while (!stack.isEmpty()) {
                const element = stack.pop();
                if (left.includes(element) && matchStr !== element)  return false;
                if (element === matchStr) break;
            }
        }
    }
    return stack.isEmpty();
};
class Stack {
    constructor() {
        this.count = 0;
        this.items = [];
    }
    push(element) {
        this.items[this.count] = element;
        this.count++;
    }
    pop() {
        if (this.isEmpty()) return undefined;
        const element = this.items[this.count - 1];
        delete this.items[this.count - 1];
        this.count--;
        return element;
    }
    isEmpty() {
        return this.size() === 0;
    }
    size() {
        return this.count;
    }
}

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简化路径【栈】

👉 【LeetCode 直通车】：71 简化路径（中等）

题解

→点击展开查看


/**
 * @param {string} path
 * @return {string}
 */
var simplifyPath = function(path) {
    let newPath = path.split('/');
    newPath = newPath.filter(item => item !== "");
    const stack = new Stack();
    for (let s of newPath) {
        if (s === '..') stack.pop();
        else if (s !== '.') stack.push(s);
    }
    if (stack.isEmpty()) return '/';
    let str = '';
    while (!stack.isEmpty()) {
        const element = stack.pop();
        str = '/' + element + str;
    }
    return str;
};
function handleBack(stack, tag, num) {
    if (!stack.isEmpty()) return num;
    const element = stack.pop();
    if (element === '..') return handleBack(stack, tag, num + 1);
    else {
        stack.push(element);
        return num;
    }
}
class Stack {
    constructor() {
        this.count = 0;
        this.items = [];
    }
    push(element) {
        this.items[this.count] = element;
        this.count++;
    }
    pop() {
        if (this.isEmpty()) return undefined;
        const element = this.items[this.count - 1];
        delete this.items[this.count - 1];
        this.count--;
        return element;
    }
    size() {
        return this.count;
    }
    isEmpty() {
        return this.size() === 0;
    }
}

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【🔥】高频算法题系列：DFS

主要有以下几类高频考题：

岛屿的最大面积【中等】【DFS】
相同的树【简单】【DFS】

岛屿的最大面积【DFS】

👉 【LeetCode 直通车】：695 岛屿的最大面积（中等）

题解

→点击展开查看


/**
 * @param {number[][]} grid
 * @return {number}
 */
let maxX, maxY;let visited;let globalMaxArea;
var maxAreaOfIsland = function(grid) {
    visited = new Set();
    maxX = grid.length;
    maxY = grid[0].length;
    globalMaxArea = 0;
    for (let i = 0; i < maxX; i++) {
        for (let j = 0; j < maxY; j++) {
            if (grid[i][j] === 1) {
                visited.add(`(${i}, ${j})`);
                globalMaxArea = Math.max(globalMaxArea, dfs(grid, i, j));
            }
            visited.clear();
        }
    }
    return globalMaxArea;
};
function dfs(grid, x, y) {
    let res = 1;
    for (let i = -1; i <= 1; i++) {
        for (let j = -1; j <= 1; j++) {
            if (Math.abs(i) === Math.abs(j)) continue;
            const newX = x + i;
            const newY = y + j;
            if (newX >= maxX || newX < 0 || newY >= maxY || newY < 0) continue;
            if (visited.has(`(${newX}, ${newY})`)) continue;
            visited.add(`(${newX}, ${newY})`);
            const areaCnt = grid[newX][newY]
            if (areaCnt === 1) {
                const cnt = dfs(grid, newX, newY);
                res += cnt;
            } 
        }
    }
    return res;
}

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相同的树【DFS】

👉 【LeetCode 直通车】：100 相同的树（简单）

题解

→点击展开查看


/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {boolean}
 */
var isSameTree = function(p, q) {
    if (p == null && q == null) return true;
    if (p == null || q == null) return false;
    if (p.val !== q.val) return false;
    return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
};

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【🔥】高频算法题系列：回溯算法

主要有以下几类高频考题：

N皇后【困难】【回溯算法】【面试真题】
全排列【中等】【回溯算法】
括号生成【中等】【回溯算法】
复原 IP 地址【中等】【回溯算法】
子集【简单】【回溯算法】

【面试真题】N皇后【回溯算法】

👉 【LeetCode 直通车】：51 N皇后（困难）

题解

→点击展开查看


/**
 * @param {number} n
 * @return {string[][]}
 */
let result = [];
var solveNQueens = function(n) {
    result = [];
    let board = [];
    for (let i = 0; i < n; i++) {
      board[i] = [];
      for (let j = 0; j < n; j++) {
        board[i][j] = '.'
      }
    }
    backtrack(0, board, n);
    return result;
};
function deepClone(board) {
  let res = [];
  for (let i = 0; i < board.length; i++) {
    res.push(board[i].join(''));
  }
  return res;
}
function backtrack(row, board, n) {
    if (row === n) {
      result.push(deepClone(board));
      return;
    }
    for (let j = 0; j < n; j++) {
        if (checkInValid(board, row, j, n)) continue;
        board[row][j] = 'Q';
        backtrack(row + 1, board, n);
        board[row][j] = '.';
      }
}
function checkInValid(board, row, column, n) {
  // 行
  for (let i = 0; i < n; i++) {
    if (board[i][column] === 'Q') return true;
  }
  for (let i = row - 1, j = column + 1; i >= 0 && j < n; i--, j++) {
    if (board[i][j] === 'Q') return true;
  }
  for (let i = row - 1, j = column - 1; i >= 0 && j >= 0; i--, j--) {
    if (board[i][j] === 'Q') return true;
  }
  return false;
}

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全排列【回溯算法】

👉 【LeetCode 直通车】：46 全排列（中等）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {number[][]}
 */
let results = [];var permute = function(nums) {
    results = [];
    backtrack(nums, []);
    return results;
};
function backtrack(nums, track) {
    if (nums.length === track.length) {
        results.push(track.slice());
        return;
    }
    for (let i = 0; i < nums.length; i++) {
        if (track.includes(nums[i])) continue;
        track.push(nums[i]);
        backtrack(nums, track);
        track.pop();
    }
}

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括号生成【回溯算法】

👉 【LeetCode 直通车】：22 括号生成（中等）

题解

→点击展开查看


/**
 * @param {number} n
 * @return {string[]}
 */
var generateParenthesis = function(n) {
    let validRes = [];
    backtrack(n * 2, validRes, '');
    return validRes;
};
function backtrack(len, validRes, bracket) {
    if (bracket.length === len) {
        if (isValidCombination(bracket)) {
            validRes.push(bracket);
        }
        return;
    }
    for (let str of ['(', ')']) {
        bracket += str;
        backtrack(len, validRes, bracket);
        bracket = bracket.slice(0, bracket.length - 1);
    }
}
function isValidCombination(bracket) {
    let stack = new Stack();
    for (let i = 0; i < bracket.length; i++) {
        const str = bracket[i];
        if (str === '(') {
            stack.push(str);
        } else if (str === ')') {
            const top = stack.pop();
            if (top !== '(') return false;
        }
    }
    return stack.isEmpty();
}
class Stack {
    constructor() {
        this.count = 0;
        this.items = [];
    }
    push(element) {
        this.items[this.count] = element;
        this.count++;
    }
    pop() {
        if (this.isEmpty()) return;
        const element = this.items[this.count - 1];
        delete this.items[this.count - 1];
        this.count--;
        return element;
    }
    size() {
        return this.count;
    }
    isEmpty() {
        return this.size() === 0;
    }
}

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复原 IP 地址【回溯算法】

👉 【LeetCode 直通车】：93 复原 IP 地址（中等）

题解

→点击展开查看


/**
 * @param {string} s
 * @return {string[]}
 */
var restoreIpAddresses = function(s) {
    if (s.length > 12) return [];
    let res = [];
    const track = [];
    backtrack(s, track, res);
    return res;
};
function backtrack(s, track, res) {
    if (track.length === 4 && s.length === 0) {
        res.push(track.join('.'));
        return;
    }
    let len = s.length >= 3 ? 3 : s.length;
    for (let i = 0; i < len; i++) {
        const c = s.slice(0, i + 1);
        if (parseInt(c) > 255) continue;
        if (i >= 1 &&  parseInt(c) < parseInt((1 + '0'.repeat(i)))) continue;
        track.push(c);
        backtrack(s.slice(i + 1), track, res);
        track.pop();
    }
}

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子集【回溯算法】

👉 【LeetCode 直通车】：78 子集（中等）

题解

→点击展开查看


/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var subsets = function(nums) {
    if (nums.length === 0) return [[]];
    let resArr = [];
    backtrack(nums, 0, [], resArr);
    return resArr;
};
function backtrack(nums, index, subArr, resArr) {
    if (Array.isArray(subArr)) {
        resArr.push(subArr.slice());
    }
    if (index === nums.length) {
        return;
    } 
    for (let i = index; i < nums.length; i++) {
        subArr.push(nums[i]);
        backtrack(nums, i + 1, subArr, resArr);
        subArr.pop(nums[i]);
    }
}

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原文地址干货文章分享 - 字节跳动最常考的 64 道JS算法题

原创不易

喜欢的话原创不易，给点鼓励吧 ❤️ 别忘了分享、点赞、在看三连哦~。

程序员财富自由之路自媒体篇 | 3000字干货分享

fengmao — Tue, 14 Jul 2020 07:43:45 +0000

疫情期间，很多人的收入多多少少都受到了影响。当然，有的人坐吃山空，也有的人收入翻倍。

我仔细观察了一下这些收入翻倍的人，发现他们基本都离不开三个字。哪三个字？就是自媒体。

自媒体是什么？

自媒体是指以内容为中心，个人创作者以网络的形式进行内容传播，从而获得一定关注度和影响力的方式。自媒体的内容形态非常广泛，文章、动态短文、音频、闻达、图集、短视频、长视频、直播等等都是自媒体可以采用的形式。

目前从事自媒体的人群有多么庞大呢？根据艾瑞网的调查数据，目前我国全职从事自媒体的人数达到了370万以上，而兼职人群则超过了600万。可见，做自媒体已经是大势所趋。

自媒体做得好的人，到底有多赚？

精品大号的“一条”，非头条广告费就达到了百万；

家庭美学定位的“卡娃微卡”，一条广告费超过10万；

papi酱，在微博上迅速走红后，短期内广告价格上涨到了2200万……

（微信公众号平台报价）

（抖音星图达人的预估报价）

博多·舍费尔在《财富自由之路》中分析个人收入构成时提到，个人收入的高低其实与五个因素息息相关，分别是“能力、精力、影响力/知名度、创意和自我评价”，而其中，影响力是最强的杠杆。

自媒体就是你能够利用好的扩大个人影响力的最佳方式。抓住了自媒体机会的人，现在都已经赚得盆满钵满。

如果你只是低头做事，那么你的才华很有可能只有老板知道，也只有老板会付给你薪酬。但是，当你在自媒体上做分享时，你的知识可以送达千千万万的粉丝，他们同样会为你的付出买单。你只需要工作一次，就能获得上不封顶的报酬。

一份时间卖出一次，想拿到百万年薪难度系数太高。一份时间卖出很多次，实现财富自由，其实并没有想象中那么难。

或许你会问，普通人也能通过做自媒体赚到钱吗？

答案是肯定的。农村四哥王荣琪，是一名从四川泸州到南方的打工仔，他从零开始经营今日头条，短短几个月，就拥有了380万粉丝，收益过万，还获得“2019今日头条年度生机创作者”称号。

（农村四哥的演讲）

现在的自媒体风格非常包容，哪怕你只是一个普通人，只要你掌握了一定的方法，做的内容有趣、有料，那么，你就能收获粉丝的信赖，在自媒体的红利中分得一杯羹。

凯文·凯利的1000粉丝理论告诉我们，只要能找到1000个铁杆粉丝，他们愿意为你的内容买单，你就足以生存得很好。

现在能赚钱的自媒体平台有哪些呢？

这个可就多了。图文类的自媒体平台有：今日头条、百家号、企鹅号、大鱼号、网易号、简书、搜狐视频号、微信公众号、知乎等等。

视频类的自媒体平台有：抖音小视频、西瓜视频、火山小视频、快手等等。

不管你是想做哪个类型内容，都不愁找不到平台发布。而且，不同的平台对于创作者都还会有相应的扶持计划和奖励计划。只要你持续产出高质量内容，就不愁没有收入。

下面详细地为大家讲解三个目前最能赚钱的平台：

1、今日头条

平台特点：

今日头条是北京字节跳动有限公司开发的一款搜索引擎产品，特点在于算法推荐，自动匹配内容和用户。新手想要从自媒体赚到钱，从今日头条入手是最好的，因为今日头条是官方鼓励赚钱，平台流量非常大，日活1.2亿+，只要你坚持发布内容就能找到对应的粉丝。

运营方法：

今日头条可以通过微头条、图集、长图文、问答等内容形式来变现，收益十分可观。

微头条：非常自由，字数、配图、领域、格式都没有限制，涨粉快，达到一定粉丝数量还可以开通商品带货功能，轻轻松松就能获得卖货收益。通过写优质微头条，可以获得加V认证，即“XX 领域优质创作者”勋章，后续能获得更多的流量扶持。

图文内容：图文内容也有阅读收益，如果拿到原创标，收益会更高。针对优质的长图文，今日头条推出青云计划，单篇图文不仅仅可以获得阅读收益，还能获得平台奖励，最低收益300元，上不封顶。

问答变现：今日头条的问答有另一个名称，叫悟空问答。只要开通了问答收益，针对你感兴趣或者有把握的领域进行回答，就能通过回答问题获得一定的收益。问答里面也有“问答青云计划”，一档奖金300，二档奖金100，随手一个回答就也能有三位数的收入，不香么？

2、百家号

平台特点：

百家号是百度专门为内容创作者打造的一个集内容创作、发布及变现为一体的互联网平台。背靠百度这个最大的中文搜索引擎网站，百家号的流量和收益也是值得关注的。

运营方法：

图文收益：百家号的图文收益目前是10000阅读/30元，可以说是主流自媒体平台之首了。在百家号上写图文内容，只要阅读量过万，获得几百收益并不难。

其他收益：百家号和今日头条相似，同样也有广告分成，问答收入，原创标收益，并且可以自荐文章，如果入选的话就可以获得至少100元+的收入和流量支持。

带货变现：百家号的文章中可以插入商品链接，通过写文章来带货，获得一定的带货分成。有的人通过带货，一篇文章轻轻松松就能获得几千收益。

同时，百家号上面也有非常多的创作者扶持计划，例如“百+计划”，百+计划的上榜者最高可以获得2万元的收益。很多的创作活动也都可以参与，都有机会获得金额不等的奖励。

3、微信公众号

平台特点：

微信公众号是基于微信社交平台而存在的个人自媒体，其重要性和必要性不言而喻。虽然说，微信公众号现在已经是大号林立，但是也依然有很多自媒体人，独辟蹊径，后来居上，短时间内获得大量粉丝，成为一匹突围的黑马。现在微信也相应开放了视频号，微信公众号极有可能会迎来下一个春天。

运营方法：

相对于今日头条和百家号，微信公众号对于自媒体人的要求会稍微高一些。因为微信公众号一般是需要通过爆款文章、刷屏文章来获得关注，粉丝数和粉丝忠诚度决定了微信公众号能够赚到多少钱。

微信公众号的收益主要来自于硬广和软广。微信公众号图文下面会投放广告，通过粉丝点击广告而获得收入，但这个收入不是特别多。真正可观的是微信公众号号主自己接广告，发软文，一般来说，一条软文广告的收益在几千到百万不等。

另外，微信公众号号主自身也可以在公众号上推出付费阅读、社群、课程等付费服务，从而获得一定的收益。

做自媒体需要具备哪些技能？如何培养这些技能？

1、写作能力

做自媒体的人都需要一定的文字功底，这是最基础的能力。不要求你妙笔生花，文采斐然，但是最起码你要把内容讲清楚，讲透彻。首先，你要能够把自己的观点完整地叙述出来，思路顺畅，然后在此基础上去追求文字的技巧性，做到把握节奏，引人入胜，让人有读下去的欲望。

对于写作的把握和对文字的驾驭，这些都是可以通过技能学习和长期练习得来的。对于文案的写作，其实都有一定的规律可循，爆款文章的思路很多时候都是共通的。只要你懂得了其中的思维，自然就可以举一反三。

2、用户思维

毫无疑问，做自媒体的人，粉丝就是我们的经济来源。如果我们总是按照我们自己的想法去做自媒体，写自己想写的文章，那么后果就是阅读量惨淡，折腾一个月也不见得有收益。真正想赚到钱，我们就应该拥有用户思维，去关注用户的关注点在哪里，用户的真正需求是什么，从而做好自己的账号定位和内容规划。

用户思维要怎么培养？第一步，对标。找到自己所在领域自媒体排行榜前10的大号。第二部，拆解。去做账号的拆解和分析，剖析他们的账号定位、选题方向、文章风格、粉丝运营都是怎么做的。在能力允许的情况下，还可以开展市场调研，真正了解你的目标用户需求，再对症下药，给出解决方案。

3、紧跟热点

现在的信息爆炸社会，每个人每天都要应对成百上千的信息资讯。你的内容如何在众多的内容中脱颖而出，就取决于你的内容是否能够第一时间吸引到用户的注意力。

眼球经济时代，顾名思义，就是你的内容必须第一时间抓住用户的眼球，没有注意力就没有利润，没有利润就注定要失败，而获取用户注意力的捷径就是紧跟热点。

何为热点？其实是用户自己投票选出来的议题。也就是说，不需要我们再去苦心积虑地做选题，只要紧跟热点，这些内容就是用户自己想要看到的。例如，最近乘风破浪的姐姐们十分火爆，只要与该热点相关的内容，基本上阅读量都不会太差。

所以，自媒体人一定要学会追踪热点，要有把自己的独家内容与时事热点相结合的能力，这样才能在保证最大打开率的同时输出自己的观点。

4、善用工具

作为自媒体人，一个人同时要兼顾许多工作，既要收集资料，又要做好选题，既要撰写文章，还要会P图修图，常常会手忙脚乱，费时又费力。如果能善用工具，就能起到事半功倍的效果。

那么，自媒体人要怎么培养这个技能呢？首先就是要时刻名铭记工具思维，当遇到一个任务的时候，可以搜索一下是否有工具可以简化该任务流程，或者可以提高工作效率，对于别人推荐的工具，可以多进行尝试，选择适合自己的工具。

在这里介绍几个自媒体人经常会用到的工具：

1）资料收集方面：

各类榜单排名大全，绝对是选题蹭热点的不二利器

大数据导航网http://hao.199it.com/

涵盖了各类指数网站和互联网工具，非常适合寻找相关材料和论据。

2）文章撰写方面：

石墨文档：https://shimo.im/welcome

实时保存，多人协作，手机和电脑均可操作，方便快捷，不用担心用Word中途断电的情况

坚果云：https://www.jianguoyun.com/

可以实现多终端保存文件，并且自动更新，手机和电脑能够同步，对于写作者来说是一大福音，随时随地都能码字了

3）文章排版方面：

秀米：https://xiumi.us/#/

一键排版、文章布局调整、团队文章等功能非常实用

135排版器：https://www.135editor.com/

样式繁多，小清新风格也是一大特色

5、图片制作网站：

创客贴：https://www.chuangkit.com/

在线图片编辑器，很多免费模板可以直接套用，封面图海报分分钟搞定

Canva：https://www.canva.cn/

简单好用的在线设计平台，设计范围更广，模板样式更偏向于高端时尚风格

由于本文的目标读者都是程序员群体，最后给大家推荐一个阅读英文技术文档的网站，https://doc.liusha.io，要想学好技术，英语是非常重要了，除了发展自媒体副业，程序员本业也是要好好发展的，大家觉得呢？

爬虫黑科技，我是怎么爬取indeed的职位数据的

fengmao — Fri, 20 Mar 2020 07:21:23 +0000

最近在学习nodejs爬虫技术，学了request模块，所以想着写一个自己的爬虫项目，研究了半天，最后选定indeed作为目标网站，通过爬取indeed的职位数据，然后开发一个自己的职位搜索引擎，目前已经上线了，虽然功能还是比较简单，但还是贴一下网址job search engine，证明一下这个爬虫项目是有用的。下面就来讲讲整个爬虫的思路。

确定入口页面

众所周知，爬虫是需要入口页面的，通过入口页面，不断的爬取链接，最后爬取完整个网站。在这个第一步的时候，就遇到了困难，一般来说都是选取首页和列表页作为入口页面的，但是indeed的列表页面做了限制，不能爬取完整的列表，顶多只能抓取前100页，但是这没有难倒我，我发现indeed有一个Browse Jobs 页面，通过这个页面，可以获取indeed按地区搜索和按类型搜索的所有列表。下面贴一下这个页面的解析代码。

start: async (page) => {
  const host = URL.parse(page.url).hostname;
  const tasks = [];
  try {
    const $ = cheerio.load(iconv.decode(page.con, 'utf-8'), { decodeEntities: false });
    $('#states > tbody > tr > td > a').each((i, ele) => {
      const url = URL.resolve(page.url, $(ele).attr('href'));
      tasks.push({ _id: md5(url), type: 'city', host, url, done: 0, name: $(ele).text() });
    });
    $('#categories > tbody > tr > td > a').each((i, ele) => {
      const url = URL.resolve(page.url, $(ele).attr('href'));
      tasks.push({ _id: md5(url), type: 'category', host, url, done: 0, name: $(ele).text() });
    });
    const res = await global.com.task.insertMany(tasks, { ordered: false }).catch(() => {});
    res && console.log(`${host}-start insert ${res.insertedCount} from ${tasks.length} tasks`);
    return 1;
  } catch (err) {
    console.error(`${host}-start parse ${page.url} ${err}`);
    return 0;
  }
}

通过cheerio解析html内容，把按地区搜索和按类型搜索链接插入到数据库中。

爬虫架构

这里简单讲一下我的爬虫架构思路，数据库选用mongodb。每一个待爬取的页面存一条记录page，包含id,url,done,type,host等字段，id用md5(url)生成，避免重复。每一个type有一个对应的html内容解析方法，主要的业务逻辑都集中在这些解析方法里面，上面贴出来的代码就是例子。

爬取html采用request模块，进行了简单的封装，把callback封装成promise，方便使用async和await方式调用，代码如下。

const req = require('request');

const request = req.defaults({
  headers: {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/78.0.3904.108 Safari/537.36'
  },
  timeout: 30000,
  encoding: null
});

const fetch = (url) => new Promise((resolve) => {
    console.log(`down ${url} started`);
    request(encodeURI(url), (err, res, body) => {
      if (res && res.statusCode === 200) {
        console.log(`down ${url} 200`);
        resolve(body);
      } else {
        console.error(`down ${url} ${res && res.statusCode} ${err}`);
        if (res && res.statusCode) {
          resolve(res.statusCode);
        } else {
          // ESOCKETTIMEOUT 超时错误返回600
          resolve(600);
        }
      }
    });
  });

做了简单的反反爬处理，把user-agent改成电脑通用的user-agent，设置了超时时间30秒，其中encoding: null设置request直接返回buffer，而不是解析后的内容，这样的好处是如果页面是gbk或者utf-8编码，只要解析html的时候指定编码就行了，如果这里指定encoding: utf-8，则当页面编码是gbk的时候，页面内容会乱码。

request默认是回调函数形式，通过promise封装，如果成功，则返回页面内容的buffer，如果失败，则返回错误状态码，如果超时，则返回600，这些懂nodejs的应该很好理解。

完整的解析代码

const URL = require('url');
const md5 = require('md5');
const cheerio = require('cheerio');
const iconv = require('iconv-lite');

const json = (data) => {
  let res;
  try {
    res = JSON.parse(data);
  } catch (err) {
    console.error(err);
   }
  return res;
};

const rules = [
  /\/jobs\?q=.*&sort=date&start=\d+/,
  /\/jobs\?q=&l=.*&sort=date&start=\d+/
];

const fns = {

  start: async (page) => {
    const host = URL.parse(page.url).hostname;
    const tasks = [];
    try {
      const $ = cheerio.load(iconv.decode(page.con, 'utf-8'), { decodeEntities: false });
      $('#states > tbody > tr > td > a').each((i, ele) => {
        const url = URL.resolve(page.url, $(ele).attr('href'));
        tasks.push({ _id: md5(url), type: 'city', host, url, done: 0, name: $(ele).text() });
      });
      $('#categories > tbody > tr > td > a').each((i, ele) => {
        const url = URL.resolve(page.url, $(ele).attr('href'));
        tasks.push({ _id: md5(url), type: 'category', host, url, done: 0, name: $(ele).text() });
      });
      const res = await global.com.task.insertMany(tasks, { ordered: false }).catch(() => {});
      res && console.log(`${host}-start insert ${res.insertedCount} from ${tasks.length} tasks`);
      return 1;
    } catch (err) {
      console.error(`${host}-start parse ${page.url} ${err}`);
      return 0;
    }
  },

  city: async (page) => {
    const host = URL.parse(page.url).hostname;
    const tasks = [];
    const cities = [];
    try {
      const $ = cheerio.load(iconv.decode(page.con, 'utf-8'), { decodeEntities: false });
      $('#cities > tbody > tr > td > p.city > a').each((i, ele) => {
        // https://www.indeed.com/l-Charlotte,-NC-jobs.html
        let tmp = $(ele).attr('href').match(/l-(?<loc>.*)-jobs.html/u);
        if (!tmp) {
          tmp = $(ele).attr('href').match(/l=(?<loc>.*)/u);
        }
        const { loc } = tmp.groups;
        const url = `https://www.indeed.com/jobs?l=${decodeURIComponent(loc)}&sort=date`;
        tasks.push({ _id: md5(url), type: 'search', host, url, done: 0 });
        cities.push({ _id: `${$(ele).text()}_${page.name}`, parent: page.name, name: $(ele).text(), url });
      });
      let res = await global.com.city.insertMany(cities, { ordered: false }).catch(() => {});
      res && console.log(`${host}-city insert ${res.insertedCount} from ${cities.length} cities`);

      res = await global.com.task.insertMany(tasks, { ordered: false }).catch(() => {});
      res && console.log(`${host}-city insert ${res.insertedCount} from ${tasks.length} tasks`);
      return 1;
    } catch (err) {
      console.error(`${host}-city parse ${page.url} ${err}`);
      return 0;
    }
  },

  category: async (page) => {
    const host = URL.parse(page.url).hostname;
    const tasks = [];
    const categories = [];
    try {
      const $ = cheerio.load(iconv.decode(page.con, 'utf-8'), { decodeEntities: false });
      $('#titles > tbody > tr > td > p.job > a').each((i, ele) => {
        const { query } = $(ele).attr('href').match(/q-(?<query>.*)-jobs.html/u).groups;
        const url = `https://www.indeed.com/jobs?q=${decodeURIComponent(query)}&sort=date`;
        tasks.push({ _id: md5(url), type: 'search', host, url, done: 0 });
        categories.push({ _id: `${$(ele).text()}_${page.name}`, parent: page.name, name: $(ele).text(), url });
      });
      let res = await global.com.category.insertMany(categories, { ordered: false }).catch(() => {});
      res && console.log(`${host}-category insert ${res.insertedCount} from ${categories.length} categories`);

      res = await global.com.task.insertMany(tasks, { ordered: false }).catch(() => {});
      res && console.log(`${host}-category insert ${res.insertedCount} from ${tasks.length} tasks`);
      return 1;
    } catch (err) {
      console.error(`${host}-category parse ${page.url} ${err}`);
      return 0;
    }
  },

  search: async (page) => {
    const host = URL.parse(page.url).hostname;
    const tasks = [];
    const durls = [];
    try {
      const con = iconv.decode(page.con, 'utf-8');
      const $ = cheerio.load(con, { decodeEntities: false });
      const list = con.match(/jobmap\[\d+\]= {.*}/g);
      const jobmap = [];
      if (list) {
         // eslint-disable-next-line no-eval
        list.map((item) => eval(item));
      }
      for (const item of jobmap) {
        const cmplink = URL.resolve(page.url, item.cmplnk);
        const { query } = URL.parse(cmplink, true);
        let name;
        if (query.q) {
          // eslint-disable-next-line prefer-destructuring
          name = query.q.split(' #')[0].split('#')[0];
        } else {
          const tmp = cmplink.match(/q-(?<text>.*)-jobs.html/u);
          if (!tmp) {
            // eslint-disable-next-line no-continue
            continue;
          }
          const { text } = tmp.groups;
          // eslint-disable-next-line prefer-destructuring
          name = text.replace(/-/g, ' ').split(' #')[0];
        }
        const surl = `https://www.indeed.com/cmp/_cs/cmpauto?q=${name}&n=10&returnlogourls=1&returncmppageurls=1&caret=8`;
        const burl = `https://www.indeed.com/viewjob?jk=${item.jk}&from=vjs&vjs=1`;
        const durl = `https://www.indeed.com/rpc/jobdescs?jks=${item.jk}`;
        tasks.push({ _id: md5(surl), type: 'suggest', host, url: surl, done: 0 });
        tasks.push({ _id: md5(burl), type: 'brief', host, url: burl, done: 0 });
        durls.push({ _id: md5(durl), type: 'detail', host, url: durl, done: 0 });
      }
      $('a[href]').each((i, ele) => {
        const tmp = URL.resolve(page.url, $(ele).attr('href'));
        const [url] = tmp.split('#');
        const { path, hostname } = URL.parse(url);
        for (const rule of rules) {
          if (rule.test(path)) {
            if (hostname == host) {
              // tasks.push({ _id: md5(url), type: 'list', host, url: decodeURI(url), done: 0 });
            }
            break;
          }
        }
      });

      let res = await global.com.task.insertMany(tasks, { ordered: false }).catch(() => {});
      res && console.log(`${host}-search insert ${res.insertedCount} from ${tasks.length} tasks`);

      res = await global.com.task.insertMany(durls, { ordered: false }).catch(() => {});
      res && console.log(`${host}-search insert ${res.insertedCount} from ${durls.length} tasks`);

      return 1;
    } catch (err) {
      console.error(`${host}-search parse ${page.url} ${err}`);
      return 0;
    }
  },

  suggest: async (page) => {
    const host = URL.parse(page.url).hostname;
    const tasks = [];
    const companies = [];
    try {
      const con = page.con.toString('utf-8');
      const data = json(con);
      for (const item of data) {
        const id = item.overviewUrl.replace('/cmp/', '');
        const cmpurl = `https://www.indeed.com/cmp/${id}`;
        const joburl = `https://www.indeed.com/cmp/${id}/jobs?clearPrefilter=1`;
        tasks.push({ _id: md5(cmpurl), type: 'company', host, url: cmpurl, done: 0 });
        tasks.push({ _id: md5(joburl), type: 'jobs', host, url: joburl, done: 0 });
        companies.push({ _id: id, name: item.name, url: cmpurl });
      }

      let res = await global.com.company.insertMany(companies, { ordered: false }).catch(() => {});
      res && console.log(`${host}-suggest insert ${res.insertedCount} from ${companies.length} companies`);

      res = await global.com.task.insertMany(tasks, { ordered: false }).catch(() => {});
      res && console.log(`${host}-suggest insert ${res.insertedCount} from ${tasks.length} tasks`);
      return 1;
    } catch (err) {
      console.error(`${host}-suggest parse ${page.url} ${err}`);
      return 0;
    }
  },

  // list: () => {},

  jobs: async (page) => {
    const host = URL.parse(page.url).hostname;
    const tasks = [];
    const durls = [];
    try {
      const con = iconv.decode(page.con, 'utf-8');
      const tmp = con.match(/window._initialData=(?<text>.*);<\/script><script>window._sentryData/u);
      let data;
      if (tmp) {
        const { text } = tmp.groups;
        data = json(text);
        if (data.jobList && data.jobList.pagination && data.jobList.pagination.paginationLinks) {
          for (const item of data.jobList.pagination.paginationLinks) {
            // eslint-disable-next-line max-depth
            if (item.href) {
              item.href = item.href.replace(/\u002F/g, '/');
              const url = URL.resolve(page.url, decodeURI(item.href));
              tasks.push({ _id: md5(url), type: 'jobs', host, url: decodeURI(url), done: 0 });
            }
          }
        }
        if (data.jobList && data.jobList.jobs) {
          for (const job of data.jobList.jobs) {
            const burl = `https://www.indeed.com/viewjob?jk=${job.jobKey}&from=vjs&vjs=1`;
            const durl = `https://www.indeed.com/rpc/jobdescs?jks=${job.jobKey}`;
            tasks.push({ _id: md5(burl), type: 'brief', host, url: burl, done: 0 });
            durls.push({ _id: md5(durl), type: 'detail', host, url: durl, done: 0 });
          }
        }
      } else {
        console.log(`${host}-jobs ${page.url} has no _initialData`);
      }
      let res = await global.com.task.insertMany(tasks, { ordered: false }).catch(() => {});
      res && console.log(`${host}-search insert ${res.insertedCount} from ${tasks.length} tasks`);

      res = await global.com.task.insertMany(durls, { ordered: false }).catch(() => {});
      res && console.log(`${host}-search insert ${res.insertedCount} from ${durls.length} tasks`);

      return 1;
    } catch (err) {
      console.error(`${host}-jobs parse ${page.url} ${err}`);
      return 0;
    }
  },

  brief: async (page) => {
    const host = URL.parse(page.url).hostname;
    try {
      const con = page.con.toString('utf-8');
      const data = json(con);
      data.done = 0;
      data.views = 0;
      data.host = host;
      // format publish date
      if (data.vfvm && data.vfvm.jobAgeRelative) {
        const str = data.vfvm.jobAgeRelative;
        const tmp = str.split(' ');
        const [first, second] = tmp;
        if (first == 'Just' || first == 'Today') {
          data.publishDate = Date.now();
        } else {
          const num = first.replace(/\+/, '');
          if (second == 'hours') {
            const date = new Date();
            const time = date.getTime();
            // eslint-disable-next-line no-mixed-operators
            date.setTime(time - num * 60 * 60 * 1000);
            data.publishDate = date.getTime();
          } else if (second == 'days') {
            const date = new Date();
            const time = date.getTime();
            // eslint-disable-next-line no-mixed-operators
            date.setTime(time - num * 24 * 60 * 60 * 1000);
            data.publishDate = date.getTime();
          } else {
            data.publishDate = Date.now();
          }
        }
      }
      await global.com.job.updateOne({ _id: data.jobKey }, { $set: data }, { upsert: true }).catch(() => { });

      const tasks = [];
      const url = `https://www.indeed.com/jobs?l=${data.jobLocationModel.jobLocation}&sort=date`;
      tasks.push({ _id: md5(url), type: 'search', host, url, done: 0 });
      const res = await global.com.task.insertMany(tasks, { ordered: false }).catch(() => {});
      res && console.log(`${host}-brief insert ${res.insertedCount} from ${tasks.length} tasks`);
      return 1;
    } catch (err) {
      console.error(`${host}-brief parse ${page.url} ${err}`);
      return 0;
    }
  },

  detail: async (page) => {
    const host = URL.parse(page.url).hostname;
    try {
      const con = page.con.toString('utf-8');
      const data = json(con);
      const [jobKey] = Object.keys(data);
      await global.com.job.updateOne({ _id: jobKey }, { $set: { content: data[jobKey], done: 1 } }).catch(() => { });
      return 1;
    } catch (err) {
      console.error(`${host}-detail parse ${page.url} ${err}`);
      return 0;
    }
  },

  run: (page) => {
    if (page.type == 'list') {
      page.type = 'search';
    }
    const fn = fns[page.type];
    if (fn) {
      return fn(page);
    }
    console.error(`${page.url} parser not found`);
    return 0;
  }

};

module.exports = fns;

每一个解析方法都会插入一些新的链接，新的链接记录都会有一个type字段，通过type字段，可以知道新的链接的解析方法，这样就能完整解析所有的页面了。例如start方法会插入type为city和category的记录，type为city的页面记录的解析方法就是city方法，city方法里面又会插入type为search的链接，这样一直循环，直到最后的brief和detail方法分别获取职位数据的简介和详细内容。

其实爬虫最关键的就是这些html解析方法，有了这些方法，你就能获取任何想要的结构化内容了。

数据索引

这部分就很简单了，有了前面获取的结构化数据，按照elasticsearch，新建一个schema，然后写个程序定时把职位数据添加到es的索引里面就行了。因为职位详情的内容有点多，我就没有把content字段添加到索引里面了，因为太占内存了，服务器内存不够用了，>_<。

DEMO

最后还是贴上网址供大家检阅，job search engine。

哔哩搜索 - 最好用的种子搜索神器

fengmao — Tue, 04 Feb 2020 01:27:03 +0000

哔哩搜索是最快的种子搜索神器，包含了近1000w的种子数据，免费提供下载

ideras job search engine launch

fengmao — Sun, 02 Feb 2020 12:03:44 +0000

ideras.com is built with angular 8 and nodejs 12, the data is crawlered with nodejs, now ideras.com has almost 10 millions jobs on our database, you can find the latest job from many job sites.
We want to build a free job search engine with better expereince than those already exist job search sites, ideras.com is beautiful and easy to use, it's very fast and smooth.
it's a great career websites for job seekers to find new jobs.
Please let me know what you think about ideras.com.

luanch of idoras.com

fengmao — Fri, 31 Aug 2018 09:31:46 +0000

I'm glad to announce the luanch of rss reader, it's written with node.js.

How does idoras.com work?

First i collect many website's feed source, and collect feed item, then use readability api to clean the item's content, and show the content on website.

DEV Community: fengmao

史上最全的文档地址列表

step by step guide tell you how to build a website like apkmirror

字节跳动最常考的 64 道JS算法题

缘起

高频算法题系列：链表

前序遍历判断回文链表

题解1

题解2

反转链表

题解

合并K个升序链表

题解

K 个一组翻转链表

题解

环形链表

题解

排序链表

题解

相交链表

题解

【🔥】高频算法题系列：字符串

【面试真题】最长回文子串【双指针】

题解

最长公共前缀【双指针】

题解

无重复字符的最长子串【双指针】

题解

【面试真题】 最小覆盖子串【滑动窗口】

题解

【🔥】高频算法题系列：数组问题

【面试真题】俄罗斯套娃信封问题【排序+最长上升子序列】

题解

最长连续递增序列【快慢指针】

题解

最长连续序列 【哈希表】

题解

【面试真题】盛最多水的容器【哈希表】

题解

寻找两个正序数组的中位数【双指针】

题解

删除有序数组中的重复项【快慢指针】

题解

和为K的子数组【哈希表】

题解

nSum问题【哈希表】【系列】

题解

【面试真题】接雨水【暴力+备忘录优化】

题解

跳跃游戏【贪心算法】【系列】

题解

高频算法题系列：二叉树

二叉树的最近公共祖先【二叉树】

题解

二叉搜索树中的搜索【二叉树】

题解

删除二叉搜索树中的节点【二叉树】

题解

完全二叉树的节点个数【二叉树】

题解

二叉树的锯齿形层序遍历【二叉树】

题解

【🔥】高频算法题系列：排序算法

用最少数量的箭引爆气球【排序算法】

题解

合并区间【排序算法+区间问题】

题解

高频算法题系列：二分查找

寻找两个正序数组的中位数【二分查找】

题解

判断子序列【二分查找】

题解

💁 在排序数组中查找元素的第一个和最后一个位置【二分搜索】

题解

【🔥】高频算法题系列：动态规划

最长递增子序列【动态规划】

题解

【面试真题】 零钱兑换【动态规划】

题解

【面试真题】 最长公共子序列【动态规划】

【面试真题】最小覆盖子串【滑动窗口】

最长连续序列【哈希表】

【面试真题】零钱兑换【动态规划】

【面试真题】最长公共子序列【动态规划】

【系列】下一个更大元素【栈】

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