[leetcode]516. Longest Palindromic Subsequence

Analysis

完了完了 真的要找不到工作了！！！！！蒼了天了！！！—— [每天刷題並不難0.0]

Given a string s, find the longest palindromic subsequence’s length in s. You may assume that the maximum length of s is 1000.

動歸解決~
dp[i][j]表示s中從i到j的子串中最長迴文串的長度，於是狀態轉移方程為：
如果s[i]=s[j]，則dp[i][j] += dp[i+1][j-1]+2；
否則dp[i][j] = max(dp[i+1][j], dp[i][j-1])。

Implement

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int len = s.size();
        vector<vector<int>> dp(len+1, vector<int>(len+1));
        for(int i=0; i<len; i++)
            dp[i][i] = 1;
        for(int i=len-1; i>=0; i--){
            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];
    }
};