题解 | #最长公共子序列(二)#

class Solution:
    def LCS(self, s1: str, s2: str) -> str:
        m, n = len(s1), len(s2)
        # 创建一个二维数组来存储最长公共子序列的长度
        dp = [[0] * (n + 1) for _ in range(m + 1)]

        # 填充dp数组
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if s1[i - 1] == s2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

        # 如果最长公共子序列的长度为0，则返回"-1"
        if dp[m][n] == 0:
            return "-1"

        # 根据dp数组构造最长公共子序列
        lcs = ""
        i, j = m, n
        while i > 0 and j > 0:
            if s1[i - 1] == s2[j - 1]:
                lcs = s1[i - 1] + lcs
                i -= 1
                j -= 1
            elif dp[i - 1][j] > dp[i][j - 1]:
                i -= 1
            else:
                j -= 1

        return lcs