leetcode--Longest Palindromic Subsequence

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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];
    }