Merge pull request #835 from casnz1601/patch-12

Update 0040.组合总和II.md

problems/0040.组合总和II.md

@@ -296,24 +296,91 @@ class Solution {
 ```
 
 ## Python 
-```python
+**回溯+巧妙去重(省去使用used**
+```python3
 class Solution:
+    def __init__(self):
+        self.paths = []
+        self.path = []
+
     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
+        '''
+        类似于求三数之和，求四数之和，为了避免重复组合，需要提前进行数组排序
+        '''
+        self.paths.clear()
+        self.path.clear()
+        # 必须提前进行数组排序，避免重复
+        candidates.sort()
+        self.backtracking(candidates, target, 0, 0)
+        return self.paths
+
+    def backtracking(self, candidates: List[int], target: int, sum_: int, start_index: int) -> None:
+        # Base Case
+        if sum_ == target:
+            self.paths.append(self.path[:])
+            return
+        
+        # 单层递归逻辑
+        for i in range(start_index, len(candidates)):
+            # 剪枝，同39.组合总和
+            if sum_ + candidates[i] > target:
+                return
+            
+            # 跳过同一树层使用过的元素
+            if i > start_index and candidates[i] == candidates[i-1]:
+                continue
+            
+            sum_ += candidates[i]
+            self.path.append(candidates[i])
+            self.backtracking(candidates, target, sum_, i+1)
+            self.path.pop()             # 回溯，为了下一轮for loop
+            sum_ -= candidates[i]       # 回溯，为了下一轮for loop
+```
+**回溯+去重（使用used）**
+```python3
+class Solution:
+    def __init__(self):
+        self.paths = []
+        self.path = []
+        self.used = []
+
+    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
+        '''
+        类似于求三数之和，求四数之和，为了避免重复组合，需要提前进行数组排序
+        本题需要使用used，用来标记区别同一树层的元素使用重复情况：注意区分递归纵向遍历遇到的重复元素，和for循环遇到的重复元素，这两者的区别
+        '''
+        self.paths.clear()
+        self.path.clear()
+        self.usage_list = [False] * len(candidates)
+        # 必须提前进行数组排序，避免重复
+        candidates.sort()
+        self.backtracking(candidates, target, 0, 0)
+        return self.paths
+
+    def backtracking(self, candidates: List[int], target: int, sum_: int, start_index: int) -> None:
+        # Base Case
+        if sum_ == target:
+            self.paths.append(self.path[:])
+            return
+        
+        # 单层递归逻辑
+        for i in range(start_index, len(candidates)):
+            # 剪枝，同39.组合总和
+            if sum_ + candidates[i] > target:
+                return
+            
+            # 检查同一树层是否出现曾经使用过的相同元素
+            # 若数组中前后元素值相同，但前者却未被使用(used == False)，说明是for loop中的同一树层的相同元素情况
+            if i > 0 and candidates[i] == candidates[i-1] and self.usage_list[i-1] == False:
+                continue
+
+            sum_ += candidates[i]
+            self.path.append(candidates[i])
+            self.usage_list[i] = True
+            self.backtracking(candidates, target, sum_, i+1)
+            self.usage_list[i] = False  # 回溯，为了下一轮for loop
+            self.path.pop()             # 回溯，为了下一轮for loop
+            sum_ -= candidates[i]       # 回溯，为了下一轮for loop
 ```
 
 ## Go：