leetcode 1071. Greatest Common Divisor of GCD Euclidean string Strings

leetcode 2020 March 1 question daily punch
EuclideanSee https://blog.csdn.net/weixin_43973433/article/details/105007066

Title:
For strings S and T, only S = T + ... + T ( T connected to itself one or more times), we identified "T S can be divisible." Returns the longest string X, X can meet the requirements, and X can be divisible divisible str1 str2.

Example 1:
Input: str1 = "ABCABC", str2 = "ABC"
Output: "ABC"
Example 2:
Input: str1 = "ABABAB", str2 = "ABAB"
Output: "AB"
Example 3:
Input: str1 = " LEET ", str2 =" CODE "
output:" "

Tip:
. 1 <= str1.length <= 1000
. 1 <= str2.length <= 1000
str1 [I] and str2 [i] uppercase letters

Ideas:

1. If the two strings have the common denominator, then a + b == b + a
2. Two string length is equal to the largest common divisor of the two conventions segment length of the string

Code:

class Solution(object):
    def gcdOfStrings(self, str1, str2):
        """
        :type str1: str
        :type str2: str
        :rtype: str
        """
        # 如果两个字符串有最大公约数，则a+b == b+a
        if str1+str2 != str2+str1:
            return ""

        # 两字符串最大公约段的长度等于两字符串长度的最大公约数
        # 辗转相除法
        def gcd(m,n):
            return gcd(n,m%n) if n else m
        
        return str1[:gcd(len(str1),len(str2))]
                

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