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
        m, n = len(s1), len(s2)
        dp = [[0] * (1 + n) for _ in range(1 + m)]
        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])
        s= []
        i,j = m,n
        while dp[i][j] !=0:
            if dp[i][j] == dp[i-1][j]:
                i -=1
            elif dp[i][j] == dp[i][j-1]:
                j-=1
            else:
                i=i-1
                j=j-1
                s.append(s1[i])
        if s is None or s == []:
            return "-1"
        else:
            res = ''.join(s)
            return res[::-1]