leetcode:Palindromic Substrings

最新推荐文章于 2019-09-06 19:29:28 发布

原创最新推荐文章于 2019-09-06 19:29:28 发布 · 202 阅读

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本文介绍了一种算法，用于计算给定字符串中不同回文子串的数量。通过使用动态规划的方法，确保即使子串由相同的字符组成但起始或结束位置不同也视为不同的子串。文中提供了一个 C++ 类实现示例。

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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".

class Solution {
public:
    int countSubstrings(string s) {
        int ans = 0;
        vector<vector<bool>> dp(s.size(), vector<bool>(s.size(), false));
        
        for (int i=0; i<s.size(); i++) {
            dp[i][i] = true;
        }
        
        for (int i=0; i+1<s.size(); i++) {
            if (s[i] == s[i+1]) {
                dp[i][i+1] = true;
            }
        }
        ans = 2;
        
        for (int i=s.size()-3; i>=0; i--) {
            for (int gap = 2; i+gap<s.size(); gap++) {
                if (s[i] == s[i+gap] && dp[i+1][i+gap-1]) {
                    dp[i][i+gap] = true;
                    ans = max(ans, gap+1);
                }
            }
        }
        
        return ans;
    }
};