题解 | #最长公共子序列(二)#

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# longest common subsequence
# @param s1 string字符串 the string
# @param s2 string字符串 the string
# @return string字符串
#
class Solution:
    def LCS(self, s1: str, s2: str) -> str:
        # write code here
        n, m = len(s1), len(s2)
        if m == 0 or n == 0:
            return "-1"
        dp = [[""]*(m+1) for _ in range(n+1)]
        dp[0][1] = ""
        dp[1][0] = ""
        for i in range(1,n+1):
            for j in range(1,m+1):
                if s1[i-1] == s2[j-1]:
                    dp[i][j] = dp[i-1][j-1] + s1[i-1]
                else:
                    dp[i][j] = dp[i-1][j] if len(dp[i-1][j]) > len(dp[i][j-1]) else dp[i][j-1]
        return dp[n][m] if dp[n][m]!="" else "-1"