516. Longest Palindromic Subsequence_java longest palindromic subsequence-优快云博客

本文探讨了使用二维动态规划解决寻找字符串中最长回文子序列的问题，通过具体实例展示了算法的实现过程，并强调了循环方向等易错点。

516. Longest Palindromic Subsequence

方法1: 2-dimension dynamic programming
- 易错点

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".

方法1: 2-dimension dynamic programming

思路：

dp: i 到 j 区间内的longest subsequence长度
initialize: 每个区间内长度至少为1
transfer: 当s[i] == s[j]，如果间隔超过1，dp[i][j] = dp[i + 1][j - 1] + 2，否则dp[i][j] = 2；如果s[i] != s[j] ，尝试舍弃i或者j，取两者较大值。
return：dp[0][n - 1]

易错点

循环方向（重要！）：因为dp是从 i + 1, j - 1继承过来的，也就是上面和左面，所以 i 需要倒序。

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int n = s.size();
        vector<vector<int>> dp(n, vector<int>(n, 1));
        for (int i = n - 1; i >= 0; i --) {
            for (int j = i + 1; j < n; j++) {
                if (s[i] == s[j]) {
                    if (j - i == 1) {
                        dp[i][j] = 2;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1] + 2;
                    }
                }
                else {
                    dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[0].back();
    }
};