Medium
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".

講解：
Dynamic Programming | Set 12 (Longest Palindromic Subsequence)

dp[i][j]表示substring s[i, j]所包含的longest palindrome的長度。首先初始化長度為1的substring所包含的longest palindrome的長度, 也就是dp[i][i] = 1. 然后看長度為2的,長度為3的……，當(dāng)s.chatAt(i) == s.charAt(j)的時候，dp[i][j] = dp[i + 1][j - 1] + 2, 因為首尾字符串相同的話，相當(dāng)于多了2個；如果s.charAt(i) != s.charAt(j)的話，則dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]).比如字符串BABACDB的子串BABAC里面找dp[0][4]的時候，因為s.charAt(0) != s.charAt(4), 所以就等于BABA, ABAC當(dāng)中palindrome較長的, 這里都是3。

注意一下初始化的index, i是從大到小n - 1到0，j是從i + 1到n,不這么寫的話后面會很麻煩，可以記一下。

image.png

image.png

class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        int[][] dp = new int[n][n];
        //substring of length 1 has the longest palidrome substring of length 1
        for (int i = 0; i < n; i++){
            dp[i][i] = 1;
        }
        for (int i = n - 1; i >= 0; i--){
            for (int j = i + 1; j < n; j++){
                if (s.charAt(i) == s.charAt(j)){
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                }else {
                    //if s.chatAt(i) != s.charAt(j), then the longest palindrome length of s(i,j) is the max of s(i+1,j) ,s(i,j-1)
                    dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[0][n - 1];
    }
}