LeetCode-Valid Palindrome II

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:

• The string will only contain lowercase characters a-z. The maximum length of the string is 50000.

题意：判断一个字符串是否为回文，最多允许删除字符串中的一个字符；

解法：对于字符串来说，如果去掉首尾已经相等的字符得到的字符串为s；即s[i] != s[j]（0<=i…<= j < s.length()）；那么我们就只需要判断字符串i与j之间的是否为回文，因为允许删除一个字符，所以需要判断s.substring(i, j)和s.substring(i + 1, j + 1)是否为回文；

Java

class Solution {
    public boolean validPalindrome(String s) {
        int begin = 0;
        int end = s.length() - 1;
        while (begin < s.length() && end >= 0 && s.charAt(begin) == s.charAt(end)) {
            begin++;
            end--;
        }
        return isPalindrome(s, begin, end - 1) || isPalindrome(s, begin + 1, end);
    }
    
    //判断字符串是否是回文
    private boolean isPalindrome(String S, int begin, int end) {
        while (begin < end) {
            if (S.charAt(begin) != S.charAt(end)) return false;
            begin++;
            end--;
        }
        return true;
    }
}