From ea88604bfdf249d686e1aba4a26795a35a57dcc8 Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Tue, 1 Jun 2021 02:44:13 -0700
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 | 49 ++++++++++++++++++++++++++
 1 file changed, 49 insertions(+)

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 d9dfc62c..51a903ef 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -184,6 +184,55 @@ class Solution {
 
 Python：
 
+这里使用了前缀表统一减一的实现方式
+
+```python
+class Solution:
+    def repeatedSubstringPattern(self, s: str) -> bool:  
+        if len(s) == 0:
+            return False
+        nxt = [0] * len(s)
+        self.getNext(nxt, s)
+        if nxt[-1] != -1 and len(s) % (len(s) - (nxt[-1] + 1)) == 0:
+            return True
+        return False
+    
+    def getNext(self, nxt, s):
+        nxt[0] = -1
+        j = -1
+        for i in range(1, len(s)):
+            while j >= 0 and s[i] != s[j+1]:
+                j = nxt[j]
+            if s[i] == s[j+1]:
+                j += 1
+            nxt[i] = j
+        return nxt
+```
+
+前缀表（不减一）的代码实现
+
+```python
+class Solution:
+    def repeatedSubstringPattern(self, s: str) -> bool:  
+        if len(s) == 0:
+            return False
+        nxt = [0] * len(s)
+        self.getNext(nxt, s)
+        if nxt[-1] != 0 and len(s) % (len(s) - nxt[-1]) == 0:
+            return True
+        return False
+    
+    def getNext(self, nxt, s):
+        nxt[0] = 0
+        j = 0
+        for i in range(1, len(s)):
+            while j > 0 and s[i] != s[j]:
+                j = nxt[j - 1]
+            if s[i] == s[j]:
+                j += 1
+            nxt[i] = j
+        return nxt
+```
 
 Go：