leetcode 647. Palindromic Substrings 回文子串的数量 + 动态规划DP

清疚 2022-06-03 06:50 191阅读 0赞

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.

这道题建议和[leetcode 5. Longest Palindromic Substring 最长回文子串的查找 + 按照length做DP][leetcode 5. Longest Palindromic Substring _ _ _length_DP] 和 [leetcode 516. Longest Palindromic Subsequence 最长回文子序列 + DP动态规划][leetcode 516. Longest Palindromic Subsequence _ _ DP] 和[leetcode 718. Maximum Length of Repeated Subarray 最长公共子串 + 动态规划DP][leetcode 718. Maximum Length of Repeated Subarray _ _ _DP]一起学习

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    class Solution 
    {
    public:
        int countSubstrings(string s) 
        {
            int n = s.length();
            vector<vector<bool>> dp(n, vector<bool>(n));
            for(int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    if (i >= j)
                        dp[i][j] = true;
                    else
                        dp[i][j] = false;
                }
            }
    
            int count = 0;
            for (int len = 1; len <= n; len++)
            {
                for (int i = 0; i + len < n; i++)
                {
                    int j = i + len;
                    if (s[i] == s[j])
                        dp[i][j] = dp[i + 1][j - 1];
                    else
                        dp[i][j] = false;
                }
            }
    
            for (int i = 0; i < n; i++)
            {
                for (int j = i; j < n; j++)
                {
                    if (dp[i][j] == true)
                        count++;
                }
            }
            return count;
        }
    };

[leetcode 5. Longest Palindromic Substring _ _ _length_DP]: http://blog.csdn.net/jackzhang_123/article/details/77712863
[leetcode 516. Longest Palindromic Subsequence _ _ DP]: http://blog.csdn.net/jackzhang_123/article/details/78820016
[leetcode 718. Maximum Length of Repeated Subarray _ _ _DP]: http://blo.csdn.net/jackzhang_123/article/details/78885152