题解 | #最长公共子串#

#
# 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 substring
# @param str1 string字符串 the string
# @param str2 string字符串 the string
# @return string字符串
#
class Solution:
    def LCS(self , str1: str, str2: str) -> str:
        n = len(str1)
        m = len(str2)
        dp = [[0]*(m+1) for _ in range(n+1)]
        ans = [0,0]
        for i in range(n):
            for j in range(m):
                if str1[i] == str2[j]:
                    dp[i+1][j+1] = dp[i][j]+1
                    if dp[i+1][j+1] > ans[1]:
                        ans = [i,dp[i+1][j+1]]
        return str1[ans[0]-ans[1]+1:ans[0]+1]