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]*(n+1) for _ in range(m+1)]#dp[i][j]代表s1前i字符与s2前j字符最大公共子序列长度
        for i in range(1,m+1):#初始化构建最长公共子序列长度数组，
            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][j-1],dp[i-1][j])
        ans, i, j = '', m, n
        while i>0 and j>0:#根据最长公共子序列数组和长度，从后往前寻找最长公共子序列
            if s1[i-1]==s2[j-1]:
                ans += s2[j-1]
                i -= 1
                j -= 1
            elif dp[i][j-1]>dp[i-1][j]:
                j -= 1
            else:
                i -= 1
        return -1 if not ans else ans[::-1]