125. 验证回文串

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125. 验证回文串

官方地址

https://leetcode-cn.com/problems/valid-palindrome/

给定一个字符串，验证它是否是回文串，只考虑字母和数字字符，可以忽略字母的大小写。 说明：本题中，我们将空字符串定义为有效的回文串。

示例 1:

输入: "A man, a plan, a canal: Panama"
输出: true

示例 2:

输入: "race a car"
输出: false

解法一 自顶向下

class Solution {
    public boolean isPalindrome(String s) {

        s = filterNonNumbersAndCharacters(s);
        return reverse(s).equalsIgnoreCase(s);
    }

    private String filterNonNumbersAndCharacters(String s) {
        return s.replaceAll("[^A-Za-z0-9]", "");
    }

    private String reverse(String s) {
        return new StringBuilder(s).reverse().toString();
    }
}

解法二

class Solution {
    public boolean isPalindrome(String s) {

        int left = 0, right = s.length() - 1;
        while (left < right) {
            while (left < right && !Character.isLetterOrDigit(s.charAt(left))) left++;
            while (left < right && !Character.isLetterOrDigit(s.charAt(right))) right--;

            if (left < right) {
                if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) return false;
                left++;
                right--;
            }
        }
        return true;
    }
}

class Solution:
    def isPalindrome(self, s: str) -> bool:
        l, r = 0, len(s)-1
        while l < r:
            while l<r and not s[l].isalnum():
                l+=1
            while l<r and not s[r].isalnum():
                r-=1
            if s[l].lower() != s[r].lower():
                return False
            l += 1
            r -= 1
        return True