[Leetcode] 647. Palindromic Substrings 解题报告

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本文介绍了一种使用动态规划方法来计算给定字符串中所有回文子串数量的算法。通过定义dp[i][j]表示从i到j的子字符串是否为回文串，并利用该定义递归地构建解决方案。

题目：

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.

思路：

定义dp[i][j]表示从i到j构成的子字符串是不是回文串，则可以用动态规划的方法计算出所有dp值。最后统计出所有为true的dp的个数即可。算法时间复杂度和空间复杂度都是O(n^2)。

代码：

class Solution {
public:
    int countSubstrings(string s) {
        int length = s.length();
        vector<vector<bool>> dp(length, vector<bool>(length, false));
        int ret = length;
        for (int i = 0; i < length; ++i) {
            dp[i][i] = true;
        }
        for (int i = 0; i + 1 < length; ++i) {
            if (s[i] == s[i + 1]) {
                dp[i][i + 1] = true;
                ++ret;
            }
        }
        for (int len = 3; len <= length; ++len) {
            for (int start = 0; start + len <= length; ++start) {
                int end = start + len - 1;
                if (s[start] == s[end] && dp[start + 1][end - 1]) {
                    dp[start][end] = true;
                    ++ret;
                }
            }
        }
        return ret;
    }
};