算法：680. Valid Palindrome II验证回文||

Posted 2021-06-30 架构师易筋

680. Valid Palindrome II

Given a string s, return true if the s can be palindrome after deleting at most one character from it.

Example 1:

Input: s = "aba"
Output: true

Example 2:

Input: s = "abca"
Output: true
Explanation: You could delete the character 'c'.

Example 3:

Input: s = "abc"
Output: false

Constraints:

1 <= s.length <= 105
s consists of lowercase English letters.

递归解法

题目是判断回文，并且最多允许去掉一个字符串，那么就可以去掉左边，或者右边的一个字符串。
用双指针指向最左边left和最右边right，用以下循环条件 left < right:

1. left位置 == right位置，则left++，right–;
2. 如果不相等，则还有一次机会，有两种情况，要么left++，要么right–，用或条件调用自己即可。

class Solution {
    public boolean validPalindrome(String s) {
        return helper(s, 0, s.length() - 1, false);
    }
    
    public boolean helper(String s, int l, int r, boolean isLastTime) {
        while (l < r) {
            if (s.charAt(l++) == s.charAt(r--)) continue;
            if (isLastTime == true) return false;
            return helper(s, l, r+1, true) || helper(s, l - 1, r, true);
        }
        
        return true;
    }
}