leetcode--Longest Palindromic Subsequence_longest palindromic subsequence csdn-优快云博客

本文介绍了一种使用动态规划算法解决寻找字符串中最长回文子序列的问题，详细阐述了状态转移方程及其代码实现。

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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
bbbab -> 4(bbbb)

思路：动态规划
dp[i][j] 表示String(i,j)子串里子最大子序列的长度
子问题求解如下；

dp[i][i] = 1;
if String(i) == String(j) dp[i][j] = dp[i+1][j-1];
if String(i) != String(j) dp[i][j] = max(dp[i][j-1],dp[i+1][j]);
代码实现思路：
dp =
0 0 0 0 0 ->1 0 0 0 0 ->1 * 0 0 0 ->1 * * 0 0 ->1 * * * 0 ->1 * * * *
0 0 0 0 0 ->0 1 0 0 0 ->0 1 * 0 0 ->0 1 * * 0 ->0 1 * * * ->0 1 * * *
0 0 0 0 0 ->0 0 1 0 0 ->0 0 1 * 0 ->0 0 1 * * ->0 0 1 * * ->0 0 1 * *
0 0 0 0 0 ->0 0 0 1 0 ->0 0 0 1 * ->0 0 0 1 * ->0 0 0 1 * ->0 0 0 1 *
0 0 0 0 0 ->0 0 0 0 1 ->0 0 0 0 1 ->0 0 0 0 1->0 0 0 0 1 ->0 0 0 0 1

代码：

public int longestPalindromeSubseq(String s) {
        if(s.length() == 0)
            return 0;
        int[][] dp = new int[s.length()][s.length()];

        for(int i = 0;i < s.length();i++){
            for(int j = i;j < s.length();j++){
                if(i == 0)
                    dp[j-i][j] = 1;
                else if(s.charAt(j-i) == s.charAt(j))
                    dp[j-i][j] = dp[j-i+1][j-1] + 2;
                else
                    dp[j-i][j] = Integer.max(dp[j-i][j-1], dp[j-i+1][j]);
            }
        }
        return dp[0][s.length()-1];
    }