lleetcode oj java 516. Longest Palindromic Subsequence

一、问题描述：

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".

二、解决思路：

经典的动态规划的问题。假设dp[i][j] 是子串i-j之间的Longest Palindromic Subsequence。那么我们可以得到动态规划的公式：

如果 s.charAt(i) == s.charAt(j) ，说明i+1至j-1之间加上i和j形成新的 Palindromic。
    dp[i][j] = dp[i + 1][j - 1] + 2;
  否则 dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1])

初始化为1.

三、代码：

public class Solution {    public int longestPalindromeSubseq(String s) {        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];    }}