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🐱(string): 1071. 字符串的最大公因子

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`@@ -2708,4 +2708,33 @@ class Solution:`
`2708`	`2708`	`# print(i, j)`
`2709`	`2709`	`# 处理特殊情况:[1,-1,1,-1]`
`2710`	`2710`	`return res and j - i > 1`
`2711`		-```
	`2711`	+```
	`2712`	`+`
	`2713`	`+## 1071. 字符串的最大公因子`
	`2714`	`+`
	`2715`	`+[原题链接](https://leetcode-cn.com/problems/greatest-common-divisor-of-strings/)`
	`2716`	`+`
	`2717`	`+### 暴力枚举`
	`2718`	`+`
	`2719`	+最大公因子必定是 `str1` 或 `str2` 的前缀,因此枚举前缀的长度并截取前缀进行匹配即可。
	`2720`	`+`
	`2721`	+```python
	`2722`	`+class Solution:`
	`2723`	`+ def gcdOfStrings(self, str1: str, str2: str) -> str:`
	`2724`	`+ length1 = len(str1)`
	`2725`	`+ length2 = len(str2)`
	`2726`	`+ for i in range(min(length1, length2), 0, -1):`
	`2727`	`+ # 枚举`
	`2728`	`+ if length1 % i == 0 and length2 % i == 0:`
	`2729`	`+ # 是否可构成前缀`
	`2730`	`+ if str1[:i] * (length1 // i) == str1 and str1[:i] * (length2 // i) == str2:`
	`2731`	`+ return str1[:i]`
	`2732`	`+ return ''`
	`2733`	+```
	`2734`	`+`
	`2735`	`+- 时间复杂度:`
	`2736`	`+- 空间复杂度:`
	`2737`	`+`
	`2738`	`+### 数学法`
	`2739`	`+`
	`2740`	+若存在 `X`,那么有 `str1 + str2 = str2 + str1`。

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