From 8fb77f425182ad1c6ebd476eedcea6c779666d13 Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Tue, 25 May 2021 11:08:42 +0200
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添加 0040.组合总和ⅠⅠ python3版本
---
 problems/0040.组合总和II.md | 22 +++++++++++++++++++---
 1 file changed, 19 insertions(+), 3 deletions(-)

diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md
index 8ff31695..4d2996f5 100644
--- a/problems/0040.组合总和II.md
+++ b/problems/0040.组合总和II.md
@@ -292,10 +292,26 @@ class Solution {
     }
 }
 ```
-
 Python：
-
-
+```python3
+class Solution:
+    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
+        res = []
+        path = []
+        def backtrack(candidates,target,sum,startIndex):
+            if sum == target: res.append(path[:])
+            for i in range(startIndex,len(candidates)):  #要对同一树层使用过的元素进行跳过
+                if sum + candidates[i] > target: return 
+                if i > startIndex and candidates[i] == candidates[i-1]: continue  #直接用startIndex来去重,要对同一树层使用过的元素进行跳过
+                sum += candidates[i]
+                path.append(candidates[i])
+                backtrack(candidates,target,sum,i+1)  #i+1:每个数字在每个组合中只能使用一次
+                sum -= candidates[i]  #回溯
+                path.pop()  #回溯
+        candidates = sorted(candidates)  #首先把给candidates排序，让其相同的元素都挨在一起。
+        backtrack(candidates,target,0,0)
+        return res
+```
 Go：