https://leetcode.com/problems/palindromic-substrings/description/

https://www.cnblogs.com/grandyang/p/7404777.html

博客中写的<=2,实际上<=1也是可以的

相当于判断一个大指针内所有子字符串是否可能为回文

class Solution {
public:
    int countSubstrings(string s) {
        int length = s.size();
        int res = ;
        vector<vector<bool>> dp(length+,vector<bool>(length+,false));
        for(int i = ;i <= length;i++){
            for(int j = ;j <= i;j++){
                if(s[i-] == s[j-]){
                    if(i -j <=  || dp[i-][j+]){
                        dp[i][j] = true;
                        res++;
                    }
                }
            }
        }
        return res;
    }
};