From 5000a978fd41dd32a639935459365fd58e691c37 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Sat, 6 May 2023 20:05:02 -0500
Subject: [PATCH] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D=E7=9A=84?=
 =?UTF-8?q?=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md?=
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---
 problems/0459.重复的子字符串.md | 39 ++++++++++++++++++++++++--
 1 file changed, 36 insertions(+), 3 deletions(-)

diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md
index 98c02a25..5d56ad18 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -264,8 +264,7 @@ class Solution {
 
 Python：
 
-这里使用了前缀表统一减一的实现方式
-
+（版本一） 前缀表 减一
 ```python
 class Solution:
     def repeatedSubstringPattern(self, s: str) -> bool:  
@@ -289,7 +288,7 @@ class Solution:
         return nxt
 ```
 
-前缀表（不减一）的代码实现
+（版本二） 前缀表 不减一
 
 ```python
 class Solution:
@@ -314,6 +313,40 @@ class Solution:
         return nxt
 ```
 
+
+（版本三） 使用 find
+
+```python
+class Solution:
+    def repeatedSubstringPattern(self, s: str) -> bool:
+        n = len(s)
+        if n <= 1:
+            return False
+        ss = s[1:] + s[:-1] 
+        print(ss.find(s))              
+        return ss.find(s) != -1
+```
+
+（版本四） 暴力法
+
+```python
+class Solution:
+    def repeatedSubstringPattern(self, s: str) -> bool:
+        n = len(s)
+        if n <= 1:
+            return False
+        
+        substr = ""
+        for i in range(1, n//2 + 1):
+            if n % i == 0:
+                substr = s[:i]
+                if substr * (n//i) == s:
+                    return True
+                
+        return False
+```
+
+
 Go：
 
 这里使用了前缀表统一减一的实现方式