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".

第一感覺就是DP，可惜寫不出來，只想到用一維DP的思想，前i位的LPS，但總覺得這么想沒什么意義，因為跟結尾也有關啊。

看了答案：

dp[i][j]: the longest palindromic subsequence's length of substring(i, j)
State transition:
dp[i][j] = dp[i+1][j-1] + 2 if s.charAt(i) == s.charAt(j)
otherwise, dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1])
Initialization: dp[i][i] = 1

畫個圖比較清楚

public class Solution {
    public int longestPalindromeSubseq(String s) {
        int[][] dp = new int[s.length()][s.length()];
        
        for (int i = s.length() - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i+1; j < s.length(); 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][s.length()-1];
    }
}

從后往前循環是因為dp[i][j]代表從i到j的LPS。如果從前往后，就要改成dp[j][i]。