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".

下面是花哥的答案，本质上是从le = 1拓展到le = n - 1，什么是le呢？le是i和j加起来占的位置，

i是左起index，j是右边end的index，我们通过比较这两者是不是相等来扩展更新dp数组，dp数组的含义是index从i到j中的最长的palin subsequence，所以理应从长度1开始

 

class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        int[][] dp = new int[n][n];
        for(int le = 1; le <= n; le++){
            for(int i = 0; i <= n - le; i++){
                int j = i + le - 1;
                if(i == j) {
                    dp[i][j] = 1;
                    continue;
                }
                if(s.charAt(i) == s.charAt(j)) dp[i][j] = 2 + dp[i+1][j-1];
                else dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
            }
        }
        return dp[0][s.length() - 1];
    }
}

https://docs.google.com/presentation/d/1KhxVVgI8jzc-g7unDNKFiHY6XDNVSK6LNsadxB14K3U/edit#slide=id.g4dab135aa1_0_468

 

 

class Solution {
    public int longestPalindromeSubseq(String s) {
        int[][] dp = new int[s.length()][s.length()];
        for(int i = s.length() - 1; i >= 0; i--){
            dp[i][i] = 1;
            for(int j = i + 1; j < s.length(); j++){
                if(s.charAt(i) == s.charAt(j)) dp[i][j] = 2 + dp[i + 1][j - 1];
                else dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
            }
        }
        return dp[0][s.length() - 1];
    }
}

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