From b033a08c13547a43f3ac4a873f65151987569e6a Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Sat, 27 May 2023 20:53:39 -0500
Subject: [PATCH] =?UTF-8?q?Update=200046.=E5=85=A8=E6=8E=92=E5=88=97.md?=
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Content-Type: text/plain; charset=UTF-8
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---
 problems/0046.全排列.md | 71 ++++++++------------------------------
 1 file changed, 15 insertions(+), 56 deletions(-)

diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md
index e08aec94..029a98e3 100644
--- a/problems/0046.全排列.md
+++ b/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):
-        self.path = []
-        self.paths = []
+    def permute(self, nums):
+        result = []
+        self.backtracking(nums, [], [False] * len(nums), result)
+        return result
 
-    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(usage_list)
-        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[:])
+    def backtracking(self, nums, path, used, result):
+        if len(path) == len(nums):
+            result.append(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 usage_list[i] == True:
+        for i in range(len(nums)):
+            if used[i]:
                 continue
-            usage_list[i] = True
-            self.path.append(nums[i])
-            self.backtracking(nums, usage_list)     # 纵向传递使用信息，去重
-            self.path.pop()
-            usage_list[i] = False
-```
-**回溯+丢掉usage_list**
-```python
-class Solution:
-    def __init__(self):
-        self.path = []
-        self.paths = []
+            used[i] = True
+            path.append(nums[i])
+            self.backtracking(nums, path, used, result)
+            path.pop()
+            used[i] = False
 
-    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