Update 0046.全排列.md

problems/0046.全排列.md

@@ -213,68 +213,27 @@ class Solution {
 ```
 
 ### Python 
-**回溯**
+回溯 使用used
 ```python 
 class Solution:
-    def __init__(self):
+    def permute(self, nums):
-        self.path = []
+        result = []
-        self.paths = []
+        self.backtracking(nums, [], [False] * len(nums), result)
+        return result
 
-    def permute(self, nums: List[int]) -> List[List[int]]:
+    def backtracking(self, nums, path, used, result):
-        '''
+        if len(path) == len(nums):
-        因为本题排列是有序的，这意味着同一层的元素可以重复使用，但同一树枝上不能重复使用(usage_list)
+            result.append(path[:])
-        所以处理排列问题每层都需要从头搜索，故不再使用start_index
-        '''
-        usage_list = [False] * len(nums)
-        self.backtracking(nums, usage_list)
-        return self.paths
-
-    def backtracking(self, nums: List[int], usage_list: List[bool]) -> None:
-        # Base Case本题求叶子节点
-        if len(self.path) == len(nums):
-            self.paths.append(self.path[:])
             return
-
+        for i in range(len(nums)):
-        # 单层递归逻辑
+            if used[i]:
-        for i in range(0, len(nums)):  # 从头开始搜索
-            # 若遇到self.path里已收录的元素，跳过
-            if usage_list[i] == True:
                 continue
-            usage_list[i] = True
+            used[i] = True
-            self.path.append(nums[i])
+            path.append(nums[i])
-            self.backtracking(nums, usage_list)     # 纵向传递使用信息，去重
+            self.backtracking(nums, path, used, result)
-            self.path.pop()
+            path.pop()
-            usage_list[i] = False
+            used[i] = False
-```
-**回溯+丢掉usage_list**
-```python
-class Solution:
-    def __init__(self):
-        self.path = []
-        self.paths = []
 
-    def permute(self, nums: List[int]) -> List[List[int]]:
-        '''
-        因为本题排列是有序的，这意味着同一层的元素可以重复使用，但同一树枝上不能重复使用
-        所以处理排列问题每层都需要从头搜索，故不再使用start_index
-        '''
-        self.backtracking(nums)
-        return self.paths
-
-    def backtracking(self, nums: List[int]) -> None:
-        # Base Case本题求叶子节点
-        if len(self.path) == len(nums):
-            self.paths.append(self.path[:])
-            return
-
-        # 单层递归逻辑
-        for i in range(0, len(nums)):  # 从头开始搜索
-            # 若遇到self.path里已收录的元素，跳过
-            if nums[i] in self.path:
-                continue
-            self.path.append(nums[i])
-            self.backtracking(nums)
-            self.path.pop()
 ```
 
 ### Go