【LeetCode】516. Longest Palindromic Subsequence最长回文子序列（非连续）

Given a string s, find the longest palindromic subsequence’s length in s. You may assume that the maximum length of s is 1000.
Input: “bbbab”
Output: 4
一种可能的最长子序列为”bbbb”.
Input: “cbbd”
Output: 2
一种可能的最长子序列为”bb”.

---

动态规划。
dp[i][j]：表示子串s[i…j]的最长回文子序列的长度，i ≤ j，i, j = 0 … n-1
答案：dp[0][n-1]

初始化：
dp[i][i] == 1

状态转移方程：
若s[i] == s[j]，dp[i][j] = 2 + dp[i+1][j-1]，要考虑i+1=j的情况
若s[i] != s[j]，dp[i][j] = max( dp[i][j-1], dp[i+1][j] )

class Solution {public:    int longestPalindromeSubseq(string s) {        int n = s.length();        if(n < 2) return n;        //二维vector的定义，区别于二维数组        vector<vector<int>> dp(n, vector<int>(n, 1));        for(int i=0; i<n ;i++) {            dp[i][i] = 1;        }        //i<j        for(int j = 1; j < n; j++) {            for(int i = j-1; i >= 0; i--) {                if(s[i] == s[j]) {                    dp[i][j] = 2 + ( (i+1 <= j-1) ? dp[i+1][j-1] : 0);                } else {                    dp[i][j] = max(dp[i+1][j], dp[i][j-1]);                }            }        }        return dp[0][n-1];    }};