Palindrome sequence and palindrome

Longest palindromic sequence

Topic Link https://leetcode.com/problems/longest-palindromic-subsequence/

Given a string s, where to find the longest palindromic sequence. S may assume a maximum length of 1000.
The difference between the longest and the palindromic sequence palindromic a question longest substring that is a substring of a string in a continuous sequence, and the sequence is a string of characters maintain the relative position of the sequence, e.g., "bbbb" can string "bbbab" but not a sub-sequence substring.

动态规划： dp[i][j] = dp[i+1][j-1] + 2 if s.charAt(i) == s.charAt(j) otherwise, dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1])

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int len = s.size();
        if(len <= 1)
            return len;
        int dp[len+1][len+1] = {0};
        memset(dp, 0, sizeof(dp));
        for(int i=len-1; i>=0; i--) {
            dp[i][i] = 1;
            for(int j=i+1; j<len; j++) {
                if(s[i] == s[j])
                    dp[i][j] = dp[i+1][j-1] + 2;
                else 
                    dp[i][j] = max(dp[i+1][j], dp[i][j-1]);
            }
        }
        return dp[0][len-1];
    }
};

A palindromic sequence number of strings

Topic links: https://leetcode.com/problems/palindromic-substrings/

Given a string, your task is to calculate how many string palindrome substring.

Substring having a different start position or the end position, even by the same characters will be counted as a different string.

class Solution {
public:
    int countSubstrings(string s) {
        int len = s.size();
        if(len <= 1)
            return len;
        int res = 0;
        for(int i=0; i<len; i++) {
            check(s, i, i, res);
            check(s, i, i+1, res);
        }
        return res;
    }
    
    void check(string s, int i, int j, int &res) {
        while(true) {
            if(i>=0 && j < s.size() && s[i] == s[j]) {
                res ++;
                i--, j++;
            } else 
                break;
        }
        return ;
    }
};