680. Valid Palindrome II

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

双指针，字符串两边对比。
如果两边字符不相等，则更新两边指针，并分别传入辅助方法再次对比。
两个结果有一个是true则返回true。

class Solution {
    public boolean validPalindrome(String s) {
        int left = 0;
        int right = s.length() - 1;
        
        while (left < right){
            if (s.charAt(left) == s.charAt(right) ){
                left++;
                right--;
            }
            else{
                return ( checkPalindrome(s, left+1, right) || checkPalindrome(s, left, right-1));
            }
        }
        return true;
    }
    private boolean checkPalindrome(String s, int left, int right){
        while (left < right){
            if (s.charAt(left)==s.charAt(right)){
                left++;
                right--;
            }
            else{
                return false;
            }
        }
        return true;
    }    
}