2022.02.26 - SX10-53.最长回文子序列

1. 题目2. 思路

(1) 动态规划 3. 代码

dp[i][j]表示字符串s的下标为[i,j]的最长回文子序列的长度。若chars[i]==chars[j]，则dp[i][j]=dp[i+1][j-1]+2；否则，dp[i][j]取dp[i][j-1]和dp[i+1][j]中较大值。 3. 代码

public class Test {
    public static void main(String[] args) {
    }
}

class Solution {
    public int longestPalindromeSubseq(String s) {
        char[] chars = s.toCharArray();
        int n = chars.length;
        int[][] dp = new int[n][n];
        for (int i = n - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i + 1; j < n; j++) {
                if (chars[i] == chars[j]) {
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                } else {
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
                }
            }
        }
        return dp[0][n - 1];
    }
}