[leetcode-516-Longest Palindromic Subsequence]

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".

 

Example 2:
Input:

"cbbd"

Output:

2

One possible longest palindromic subsequence is "bb"

思路：

参考：

动态规划解决。设立一个len行len列的dp数组。dp[i][j]表示字符串i～j下标所构成的子串中最长回文子串的长度。最后我们需要返回的是dp[0][len-1]的值。
dp数组这样更新：首先i指针从尾到头遍历，j指针从i指针后面一个元素开始一直遍历到尾部。一开始dp[i][i]的值都为1，如果当前i和j所指元素相等，说明能够加到i～j的回文子串的长度中，所以更新dp[i][j] = dp[i+1][j-1] + 2; 如果当前元素不相等，那么说明这两个i、j所指元素对回文串无贡献，则dp[i][j]就是从dp[i+1][j]和dp[i][j-1]中选取较大的一个值即可。

https://www.liuchuo.net/archives/3204

int longestPalindromeSubseq(string s)
{
  int n = s.size();
  vector<vector<int>>dp(n,vector<int>(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];
}