[📝LeetCode #125] Valid Palindrome

#leetcode #java #100daysofcode #string

🎀 The Problem

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.

Given a string s, return true if it is a palindrome, or false otherwise.

Example:

Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.

👩‍💻 My Answer

class Solution {
    public boolean isPalindrome(String s) {

        s = s.toLowerCase();
        int i = 0;
        int j = s.length() - 1;

        while (i != j && i < s.length()) {

            while (48 > s.charAt(i) || (57 < s.charAt(i) && s.charAt(i) < 97) || s.charAt(i) > 122) {
                i++;

                if (i == s.length())
                    return true;
            }

            while (48 > s.charAt(j) || (57 < s.charAt(j) && s.charAt(j) < 97) || s.charAt(j) > 122) {
                j--;
            }

            if (s.charAt(j) != s.charAt(i))
                return false;

            i++;
            j--;
        }

        return true;
    }
}

Pro & Con

✖️ Runtime & Memory
✖️ Too long
✖️ Bit complicated (Hard to read)

💋 Ideal Answer

Approach

I read the solution post on LeetCode and realized there is a library to check if the char is a letter (Character.isLetterOrDigit())

New Code

class Solution {
    public boolean isPalindrome(String s) {

        s = s.toLowerCase();
        int i = 0;
        int j = s.length() - 1;

        while (i < j) {
            if (!Character.isLetterOrDigit(s.charAt(i))) {
                i++;
            } else if (!Character.isLetterOrDigit(s.charAt(j))) {
                j--;
            } else if (s.charAt(j) != s.charAt(i)) 
                return false;
            else {
                i++;
                j--;
            }
        }
        return true;
    }
}