Leetcode 647.Palindromic Substrings(算法分析week15)

Palindromic Substrings

-问题描述-
-算法分析-
-代码实现-

问题描述

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: 3
Explanation: Three palindromic strings: “a”, “b”, “c”.

Example 2:

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

Note:
The input string length won’t exceed 1000.

算法分析

问题大意：找到一个字符串中所有回文子串的数目。
算法分析：
1、使用动态规划解决问题。定义二维数组dp，dp[i][j]表示字符串中从i到j的字串是否为回文子串，如果是，dp[i][j] = 1；否则dp[i][j] = 0;
2、字符串中的每个元素都是一个回文子串，即dp[i][i] = 1;
3、从第i个元素到第j个元素的子串.dp[i][j] = 1当且仅当s[i]==s[j]，dp[i-1][j-1]==1或者i==j-1或者i==j-2。
4、一个字符串中所有回文子串的数目就是数组dp中所有元素值为1的数目。

代码实现

@requires_authorizationclass Solution {public:    int countSubstrings(string s) {        int sum = 0;        int n = s.length();        int **dp = new int*[n+1];        for (int i = 0; i < n; i++) {            dp[i] = new int[n+1];        }        for (int i = 0; i < n; i++) {            dp[i][i] = 1;            sum++;            for (int j = 0; j < i; j++) {                if (s[j] == s[i] && (dp[j+1][i-1] == 1||j == i - 1||j == i -2)) {                    dp[j][i] = 1;                    sum++;                }            }        }        return sum;    }};