最长回文子序列(递归+动态规划)

递归比较简单

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        return LFS(s,0,s.size()-1);    
    }
    int LFS(string s,int begin,int end) {
        if(begin==end) return 1;
        if(begin>end) return 0;
        if(s[begin]==s[end])
            return LFS(s,begin+1,end-1)+2;
        else
            return max(LFS(s,begin+1,end),LFS(s,begin,end-1));
    }
};

动态规划：dp[i][j]表示i,j之间最长回文串的长度

递推公式如下：dp[i][j]=dp[i+1][j-1]+2                       s[i]==s[j]

                         dp[i][j]=max(dp[i+1][j],  dp[i][j-1])    s[i]!=s[j]  

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