Discussion on: Daily Coding Puzzles - Oct 29th - Nov 2nd

[Ali Spittel]

Wednesday

Write a program to reverse the digits of a positive integer but without converting it to a string.

[Courtney]

probably not the best solution but it works. (unless you want leading zeros...but I'm assuming that's not the case since it wants a number).

function reverseNum(n) {
    const divisor = 10;
    let reversed = 0;
    let holder = [];
    while(n > 0) {
        holder.push(n % divisor);
        n = (Math.floor(n / divisor));
    }
    let place = Math.pow(10, holder.length-1);
    for(let i = 0; i < holder.length; i++) {
        reversed += (holder[i] * place);
        place /= 10;
    }
    return reversed;
}

[Kasey Speakman]

F#

let rec loop rev i =
    if i = 0 then rev
    else loop (rev * 10 + (i % 10)) (i / 10)

// usage, returns 987654321
let reversed = loop 0 123456789

• Uses a recursive loop. (F# will TCO to a while loop on compile)
• i % 10 gets the right-most digit of i
• rev * 10 shifts the numbers left, with right-most zero
• i / 10 shifts the numbers right, dropping right-most digit

I happened to remember these little number tricks from a previous challenge. This is basically using integers as digit stacks.

[Sung M. Kim]

Thank you, Ali.
This has been one of the most fun challenges.

Took me awhile but here is the C# version.

The gist is that, remainder is calculated for each digit, stored in a stack (LIFO - last in first out) to reverse the remainder.

Lastly the total is reconstructed from the stack.

Runnable code on .NET Fiddle

using System;
using System.Collections.Generic;

public class Program
{
    public static void Main()
    {
        var a = new int[] {123456, 5, 654321, 1234567890};

        foreach (var n in a) {
            Console.WriteLine($"Reversed = {Reverse(n)}");  
        }
    }

    private static int Reverse(int input) {
        var stack = BuildStack(input);
        return ConvertStackToNumber(stack);
    }

    private static Stack<int> BuildStack(int input) {
        var power = 0;
        var stack = new Stack<int>();

        while (true) {
            power++;
            var modBy = (int) Math.Pow(10, power);
            var divisior = (int) Math.Pow(10, power - 1);

            var remainder = (input % modBy) / divisior;
            if (remainder == 0 && power != 1) break;

            stack.Push(remainder);
        }
        return stack;
    }

    private static int ConvertStackToNumber(Stack<int> stack) {
        var total = 0;
        var power = 0;
        while (stack.Count > 0) {
            var current = stack.Pop();
            total += current * (int)Math.Pow(10, power++);
        }
        return total;
    }
}

[Ali Spittel]

Oh that's a really cool approach!

[E. Choroba]

#! /usr/bin/perl
use warnings;
use strict;

sub rev {
    my ($x) = @_;
    my $r = my $z = 0;
    my $start = 1;
    while ($x) {
        ! ($r *= 10) and $z += $start or undef $start;
        $r += $x % 10;
        $x = ($x - $x % 10) / 10;
    }
    return '0' x ($z - 1) . $r
}

use Test::More tests => 4;
is rev(3), 3;
is rev(123456789), 987654321;
is rev(4444), 4444;
is rev(1000), '0001';

[Ali Spittel]

My Python solution:

def reverse_number(num, running_value=0):
    if num == 0: 
        return running_value // 10
    quotient, remainder = divmod(num, 10)
    running_value += remainder
    return reverse_number(quotient, running_value * 10)

print(reverse_number(123456))
print(reverse_number(5))

[Sung M. Kim]

And now this is... 😮