2022.03.09 - NC045.BM65 最长公共子序列(二)

1. 题目

2. 思路

(1) 动态规划

• dp[i][j]表示s1中下标为[0,i)的子字符串与s2中下标为[0,j)的子字符串的最长公共子序列。
• 若chars[i-1]==chars[j-1]，则dp[i][j]=dp[i-1][j-1]+chars[i-1]；否则，dp[i][j]取dp[i-1][j]和dp[i][j-1]中较长的一个。

3. 代码

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

class Solution {
    /**
     * longest common subsequence
     *
     * @param s1 string字符串 the string
     * @param s2 string字符串 the string
     * @return string字符串
     */
    public String LCS(String s1, String s2) {
        char[] chars1 = s1.toCharArray();
        char[] chars2 = s2.toCharArray();
        int n1 = chars1.length;
        int n2 = chars2.length;
        String[][] dp = new String[n1 + 1][n2 + 1];
        for (int i = 0; i <= n1; i++) {
            dp[i][0] = "";
        }
        for (int i = 0; i <= n2; i++) {
            dp[0][i] = "";
        }
        for (int i = 1; i <= n1; i++) {
            for (int j = 1; j <= n2; j++) {
                if (chars1[i - 1] == chars2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + chars1[i - 1];
                } else {
                    if (dp[i - 1][j].length() > dp[i][j - 1].length()) {
                        dp[i][j] = dp[i - 1][j];
                    } else {
                        dp[i][j] = dp[i][j - 1];
                    }
                }
            }
        }
        return dp[n1][n2].equals("") ? "-1" : dp[n1][n2];
    }
}