LeetCode 进阶之路 - 125.验证回文串

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

/**
 * 125. 验证回文串
 * @Author: lixj
 * @Date: 2020/9/21 11:08
 */
public class ValidPalindrome {
    /**
     * 拼接新字符串再判断，双指针法
     * @param s
     * @return
     */
    public boolean isPalindrome(String s) {
        char[] ch = s.toCharArray();
        int len = s.length();
        if (len == 0) return true;
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i <= len -1; i++) {
            if (Character.isLetterOrDigit(ch[i])) {
                sb.append(ch[i]);
            }
        }

        int left = 0;
        int right = sb.length() -1;
        char[] chnew = sb.toString().toLowerCase().toCharArray();
        while (left < right) {
            if (chnew[left] != chnew[right]) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }

    /**
     * 原字符串判断
     * @param s
     * @return
     */
    public boolean isPalindrome1(String s) {
        int left = 0;
        int 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;
    }


    public static void main(String[] args) {
        String srt = "A man, a plan, a canal: Panama";
//        String srt = "A man, a plan, a canal: vnama";
        ValidPalindrome validPalindrome = new ValidPalindrome();
        System.out.println(validPalindrome.isPalindrome1(srt));
    }
}