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

Input: "aba"
Output: True

Input: "abca"
Output: True
Explanation: You could delete the character ‘c‘.

判斷字符串是不是回文，最多可以刪除字符串的一個元素來使它成為回文。

解決：從字符串兩端開始比較，用一個指針low從前遍歷，用high從尾部遍歷。

1、當s[low] == s[high]，說明符合回文，++low, --high，遍歷下一組。

2、要是s[low] != s[high]，就要刪除一個數，重組。重組的原則就是s[low]和s[high-1]比較，s[high]和s[low+1]比較，是否能配對。

2.1、要是s[low+1] == s[high]，那就刪除當前的s[low]；對應的s[low] == s[high-1]，就刪除當前的s[high]。

2.2、要是兩個都符合條件，就要順次看下一對能不能組合。

2.3、要是兩個都不符合條件，那就說明只刪除一個元素不能形成回文。

class Solution {
public:
    bool validPalindrome(string s) {
        int
 
 low = 0;
        int high = s.length() - 1;
        int cnt = 0;
        while (low<high) {
            if (s[low] != s[high]) {
                if (s[low+1] == s[high] && s[low] == s[high-1]) {
                    if (s[low+1] == s[high] && s[low+2] == s[high-1])
                        
 
++low;
                    else if (s[low] == s[high-1] && s[low+1] == s[high-2])
                        --high;

                }
                else if (s[low+1] == s[high])
                    ++low;
                else if (s[low] == s[high-1])
                    --high;
                else
                    return false;
                ++cnt;
            }
            if (cnt==2)
                return false;
            ++low;
            --high;
        }
        return true;
    }
};

680. Valid Palindrome II