LeetCode 516. Longest Palindromic Subsequence

原题链接在这里：https://leetcode.com/problems/longest-palindromic-subsequence/

题目：

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".

题解：

DP问题. dp[i][j]表示从i到j的最大subsequence长度.

状态转移. 如果s.charAt(i) == s.charAt(j), dp[i][j] = dp[i+1][j-1] + 2. 所以i要从大往小变化.

否则dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]). 要么去头 要么去尾, 看剩下段的最长subsequence长度.

初始化 dp[i][i] = 1.

答案., dp[0][len-1].

Time Complexity: O(n^2), n = s.length().

Space: O(n^2).

AC Java:

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

AC Python:

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        if not s:
            return 0
        n = len(s)
        dp = [[0 for _ in range(n)] for _ in range(n)]
        for i in range(n - 1, -1, -1):
            dp[i][i] = 1
            for j in range(i + 1, n):
                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]

类似Longest Palindromic Substring, Valid Palindrome III.

跟上Count Different Palindromic Subsequences.