Longest Palindromic Subsequence

方法：与substring的不同是这里不需要子串连续

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        vector<vector<int>> f(s.size(), vector<int>(s.size(), 0));

        for(int i=s.size()-1; i>=0; --i)
        {
            f[i][i] = 1;
            for(int j=i+1; j<s.size(); ++j)
            {
                if(s[i] == s[j])
                    f[i][j] = f[i+1][j-1] + 2;
                else
                    f[i][j] = max(f[i+1][j], f[i][j-1]);
            }
        }

        return f[0][s.size()-1];
    }
};

或者

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        vector<vector<int>> f(s.size(), vector<int>(s.size(), 0));

        for(int i=0; i<s.size(); ++i)
        {
            f[i][i] = 1;
            for(int j=i-1; j>=0; --j)
            {
                if(s[i] == s[j])
                    f[j][i] = f[j+1][i-1] + 2;
                else
                    f[j][i] = max(f[j+1][i], f[j][i-1]);
            }
        }

        return f[0][s.size()-1];
    }
};