LeetCode--Palindromic Substrings

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: "abc"Output: 3Explanation: Three palindromic strings: "a", "b", "c".

Example 2:

Input: "aaa"Output: 6Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

Note:

1. The input string length won't exceed 1000.

这道题的意思就是给定一个字符串，能从这个字符串的子串中找到多少个回文子串。这道题可以用动态规划的思路来解决。比如说dp[i][j]为1代表s[i]到s[j]这段子串为回文串（包括I和j），为0代表不为回文串，每次判断的时候，如果s[i]与s[j]相等，且dp[I+1][j-1]为1，那么dp[i][j]的值就是1。当i与j相差为1或者相差为2时，只要s[i]与s[j]相等，dp[i][j]的值就为1，还有就是I和j相等时dp[i][j]的值一定为1。按着这个思路，最开始从dp[0][0]的值开始入手，之后就是用动态规划的基本思路一步一步确定所有答案。（比如前三个字符组成的字串对应的dp[i][j]的值用于之后确定前四个字符组成的子串对应的dp[i][j]的值）

代码如下：

class Solution {public:    int countSubstrings(string s) {        int n = s.size(), count = 0;        vector<vector<int>> dp(n, vector<int> (n));        for (int j = 0; j < n; ++j) {            dp[j][j] = 1;            ++count;            for (int i = 0; i < j; ++i) {                if (s[i] == s[j] && (i+1 >= j-1 || dp[i+1][j-1])) {                    dp[i][j] = 1;                    ++count;                }            }        }        return count;    }};