题目描述

英文版描述

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.

英文版地址

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

中文版描述

给定一个字符串 s ，验证 s 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。本题中，将空字符串定义为有效的回文串。

示例 1: 输入: s = “A man, a plan, a canal: Panama” 输出: true 解释：”amanaplanacanalpanama” 是回文串示例 2: 输入: s = “race a car” 输出: false 解释：”raceacar” 不是回文串

Constraints·提示：

1 <= s.length <= 2 * 105
字符串 s 由 ASCII 字符组成

中文版地址

https://leetcode.cn/problems/XltzEq/https://leetcode.cn/problems/XltzEq/

解题思路

先全部转小写字母（题目要求忽略大小写）然后先去除除了字母和数字的字符，首尾依次比较

解题方法

俺这版

class Solution {
 public static boolean isPalindrome(String s) {
        String s1 = s.toLowerCase().replaceAll("[^a-z|0-9]", "");
        int length = s1.length();
        for (int i = 0; i < length; i++) {
            int j = length - i - 1;
            if(j <= i){
                break;
            }
            char c1 = s1.charAt(i);
            char c2 = s1.charAt(j);
            if (c1 != c2) {
                return false;
            }
        }
        return true;
    }
}

官方版

筛选+比较

class Solution {
        public boolean isPalindrome(String s) {
        StringBuffer sgood = new StringBuffer();
        int length = s.length();
        for (int i = 0; i < length; i++) {
            char ch = s.charAt(i);
            if (Character.isLetterOrDigit(ch)) {
                sgood.append(Character.toLowerCase(ch));
            }
        }
        StringBuffer sgood_rev = new StringBuffer(sgood).reverse();
        return sgood.toString().equals(sgood_rev.toString());
    }
}

双指针法

class Solution {
    public boolean isPalindrome(String s) {
        StringBuffer sgood = new StringBuffer();
        int length = s.length();
        for (int i = 0; i < length; i++) {
            char ch = s.charAt(i);
            if (Character.isLetterOrDigit(ch)) {
                sgood.append(Character.toLowerCase(ch));
            }
        }
        int n = sgood.length();
        int left = 0, right = n - 1;
        while (left < right) {
            if (Character.toLowerCase(sgood.charAt(left)) != Character.toLowerCase(sgood.charAt(right))) {
                return false;
            }
            ++left;
            --right;
        }
        return true;
    }
}