README_EN.md

680. Valid Palindrome II

中文文档

Description

Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.

Example 1:

Input: "aba"

Output: True

Example 2:

Input: "abca"

Output: True

Explanation: You could delete the character 'c'.

Note:

1. The string will only contain lowercase characters a-z.

   The maximum length of the string is 50000.

Solutions

Python3

class Solution:
    def validPalindrome(self, s: str) -> bool:
        def isPalindrome(s):
            i, j = 0, len(s) - 1
            while i < j:
                if s[i] != s[j]:
                    return False
                i += 1
                j -= 1
            return True

        i, j = 0, len(s) - 1
        while i < j:
            if s[i] != s[j]:
                return isPalindrome(s[i: j]) or isPalindrome(s[i + 1: j + 1])
            i += 1
            j -= 1
        return True

Java

class Solution {
    public boolean validPalindrome(String s) {
        for (int i = 0, j = s.length() - 1; i < j; ++i, --j) {
            if (s.charAt(i) != s.charAt(j)) {
                return isPalindrome(s.substring(i, j)) || isPalindrome(s.substring(i + 1, j + 1));
            }
        }
        return true;
    }

    private boolean isPalindrome(String s) {
        for (int i = 0, j = s.length() - 1; i < j; ++i, --j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
        }
        return true;
    }
}

TypeScript

function validPalindrome(s: string): boolean {
    for (let i: number = 0, j = s.length - 1; i < j; ++i, --j) {
        if (s.charAt(i) != s.charAt(j)) {
            return isPalinddrome(s.slice(i, j)) || isPalinddrome(s.slice(i + 1, j + 1));
        }
    }
    return true;
};

function isPalinddrome(s: string): boolean {
    for (let i: number = 0, j = s.length - 1; i < j; ++i, --j) {
        if (s.charAt(i) != s.charAt(j)) {
            return false;
        }
    }
    return true;
}

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