647. Palindromic Substrings

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最新推荐文章于 2025-03-25 07:15:00 发布

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分类专栏： leetcode

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本文介绍了一种通过动态规划方法高效计算字符串中所有回文子串数量的算法。给出两个示例说明如何使用该算法得出正确答案。此外，文章还提供了一个C++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".

这题比较简单，找出递推公式较为容易。dp[i] = dp[i-1] + (新增的数目).

class Solution {
public:
    bool isPalindromic(const char * start,const char *end){
        while(start <= end){
            if((*start) == (*end)){
                start++;
                end--;
            }else{
                return false;
            }
        }

        return true;
    }

    int countSubstrings(string s) {
        int n = s.size();
        vector<int> dp(n,1);

        dp[0] = 1;
        const char * start = s.c_str();
        for(int i = 1; i < n;++i){
            const char * end = start + i;
            dp[i] = dp[i-1];
            for(int j = 0; j <= i; j++){
                if(isPalindromic(start+j,end)){
                    dp[i] += 1;
                }
            }
        }

        return dp[n-1];
    }
};