Posted on Jun 20

[📝LeetCode #125] Valid Palindrome

🎀 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; } }