Update 0090.子集II.md

problems/0090.子集II.md

@@ -235,86 +235,84 @@ class Solution {
 }
 ```
 
-### Python 
-```python
-class Solution:
-    def __init__(self):
-        self.paths = []
-        self.path = []
 
-    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
-        nums.sort()
-        self.backtracking(nums, 0)
-        return self.paths
-
-    def backtracking(self, nums: List[int], start_index: int) -> None:
-        # ps.空集合仍符合要求
-        self.paths.append(self.path[:])
-        # Base Case
-        if start_index == len(nums):
-            return
-        
-        # 单层递归逻辑
-        for i in range(start_index, len(nums)):
-            if i > start_index and nums[i] == nums[i-1]:
-                # 当前后元素值相同时，跳入下一个循环，去重
-                continue
-            self.path.append(nums[i])
-            self.backtracking(nums, i+1)
-            self.path.pop()
-```
 
 #### Python3
 
-不使用used数组
+回溯 利用used数组去重
 ```python
 class Solution:
-    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
+    def subsetsWithDup(self, nums):
-        res = []
-        path = []
-        nums.sort() # 去重需要先对数组进行排序
-
-        def backtracking(nums, startIndex):
-            # 终止条件
-            res.append(path[:])
-            if startIndex == len(nums):
-                return
-            
-            # for循环
-            for i in range(startIndex, len(nums)):
-                # 数层去重
-                if i > startIndex and nums[i] == nums[i-1]: # 去重
-                    continue
-                path.append(nums[i])
-                backtracking(nums, i+1)
-                path.pop()
-        
-        backtracking(nums, 0)
-        return res
-```
-
-使用used数组
-```python
-class Solution:
-    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
         result = []
         path = []
-        nums.sort()
+        used = [False] * len(nums)
-        used = [0] * len(nums)
+        nums.sort()  # 去重需要排序
-        def backtrack(nums, startIdx):
+        self.backtracking(nums, 0, used, path, result)
-            result.append(path[:])
-            for i in range(startIdx, len(nums)):
-                if i > startIdx and nums[i] == nums[i-1] and used[i-1] == 0:
-                    continue
-                used[i] = 1
-                path.append(nums[i])
-                backtrack(nums, i+1)
-                path.pop()
-                used[i] = 0
-        backtrack(nums, 0)
         return result
+
+    def backtracking(self, nums, startIndex, used, path, result):
+        result.append(path[:])  # 收集子集
+        for i in range(startIndex, len(nums)):
+            # used[i - 1] == True，说明同一树枝 nums[i - 1] 使用过
+            # used[i - 1] == False，说明同一树层 nums[i - 1] 使用过
+            # 而我们要对同一树层使用过的元素进行跳过
+            if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]:
+                continue
+            path.append(nums[i])
+            used[i] = True
+            self.backtracking(nums, i + 1, used, path, result)
+            used[i] = False
+            path.pop()
+
 ```
 
+回溯 利用集合去重
+
+```python
+class Solution:
+    def subsetsWithDup(self, nums):
+        result = []
+        path = []
+        nums.sort()  # 去重需要排序
+        self.backtracking(nums, 0, path, result)
+        return result
+
+    def backtracking(self, nums, startIndex, path, result):
+        result.append(path[:])  # 收集子集
+        uset = set()
+        for i in range(startIndex, len(nums)):
+            if nums[i] in uset:
+                continue
+            uset.add(nums[i])
+            path.append(nums[i])
+            self.backtracking(nums, i + 1, path, result)
+            path.pop()
+
+```
+
+回溯 利用递归的时候下一个startIndex是i+1而不是0去重
+
+```python
+class Solution:
+    def subsetsWithDup(self, nums):
+        result = []
+        path = []
+        nums.sort()  # 去重需要排序
+        self.backtracking(nums, 0, path, result)
+        return result
+
+    def backtracking(self, nums, startIndex, path, result):
+        result.append(path[:])  # 收集子集
+        for i in range(startIndex, len(nums)):
+            # 而我们要对同一树层使用过的元素进行跳过
+            if i > startIndex and nums[i] == nums[i - 1]:
+                continue
+            path.append(nums[i])
+            self.backtracking(nums, i + 1, path, result)
+            path.pop()
+
+
+```
 ### Go 
 
 ```Go