From ed8ca0daae4920af17410fc4724a207cb0eedd95 Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Tue, 25 May 2021 15:14:00 +0200
Subject: [PATCH] =?UTF-8?q?Update=200131.=E5=88=86=E5=89=B2=E5=9B=9E?=
 =?UTF-8?q?=E6=96=87=E4=B8=B2.md?=
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添加 0131.分割回文串 python3版本
---
 problems/0131.分割回文串.md | 19 ++++++++++++++++++-
 1 file changed, 18 insertions(+), 1 deletion(-)

diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md
index 9b2401de..c722af37 100644
--- a/problems/0131.分割回文串.md
+++ b/problems/0131.分割回文串.md
@@ -292,7 +292,24 @@ class Solution {
 ```
 
 Python：
-
+```python3
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        res = []  
+        path = []  #放已经回文的子串
+        def backtrack(s,startIndex):
+            if startIndex >= len(s):  #如果起始位置已经大于s的大小，说明已经找到了一组分割方案了
+                return res.append(path[:])
+            for i in range(startIndex,len(s)):
+                p = s[startIndex:i+1]  #获取[startIndex,i+1]在s中的子串
+                if p == p[::-1]: path.append(p)  #是回文子串
+                else: continue  #不是回文，跳过
+                backtrack(s,i+1)  #寻找i+1为起始位置的子串
+                path.pop()  #回溯过程，弹出本次已经填在path的子串
+        backtrack(s,0)
+        return res
+                
+```
 
 Go：