Given a string s, find the longest palindromic subsequence‘s length in s. You may assume that the maximum length of s is 1000.
Example 1: Input: "bbbab" Output: 4 One possible longest palindromic subsequence is "bbbb".
判断字符串中最长的回文子串，子串不一定要连续。
动态规划的方法：
1、s[i]==s[j]，s[i,j]的最长子串就是s[i+1, j-1]能构成的回文串+2；
2、不相等，也就是s[i]，s[j]不能同时加入子串，s[i,j]的最长子串就是s[i, j-1]或s[i+1, j]能构成的回文串
1 class Solution { 2 public: 3 int longestPalindromeSubseq(string s) { 4 int len = s.length(); 5 if (len == 0) 6 return 0; 7 vector<vector<int>> dp(len, vector<int>(len)); 8 for (int i=0; i<len; ++i) 9 dp[i][i] = 1; 10 for (int i=0; i<len; ++i) 11 for (int j=i-1; j>=0; --j) { 12 if (s[i] == s[j]) { 13 dp[i][j] = dp[i-1][j+1] + 2; 14 } 15 else { 16 dp[i][j] = max(dp[i-1][j], dp[i][j+1]); 17 } 18 19 } 20 return dp[len-1][0]; 21 } 22 };
