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:
        # write code here
        re,index = -1,0
        if not str1 or not str2: return -1    
        for i in range(len(str1)):
            if str1[i-index:i+1] in str2:
                re = str1[i-index:i+1]
                index += 1
        return re