feat: 46.permutations update Python3 implementation (#152)

problems/46.permutations.md

@@ -40,10 +40,11 @@ Output:
 - 回溯法
 - backtrack 解题公式
 
-
 ## 代码
 
-* 语言支持: Javascript，Python3
+* 语言支持: Javascript, Python
+
+Javascript Code:
 
 ```js
 /*
@@ -60,10 +61,10 @@ Output:
  * Testcase Example:  '[1,2,3]'
  *
  * Given a collection of distinct integers, return all possible permutations.
- * 
+ *
  * Example:
- * 
- * 
+ *
+ *
  * Input: [1,2,3]
  * Output:
  * [
@@ -74,8 +75,8 @@ Output:
  * ⁠ [3,1,2],
  * ⁠ [3,2,1]
  * ]
- * 
- * 
+ *
+ *
  */
 function backtrack(list, tempList, nums) {
     if (tempList.length === nums.length) return list.push([...tempList]);
@@ -96,13 +97,16 @@ var permute = function(nums) {
     return list
 };
 ```
-Python3 Code:
+
+Python Code:
+
 ```Python
 class Solution:
     def permute(self, nums: List[int]) -> List[List[int]]:
         """itertools库内置了这个函数"""
+        import itertools
         return itertools.permutations(nums)
-    
+
     def permute2(self, nums: List[int]) -> List[List[int]]:
         """自己写回溯法"""
         res = []
@@ -119,6 +123,21 @@ class Solution:
                     _backtrace(left_nums, p_list)
         _backtrace(nums, [])
         return res
+
+    def permute3(self, nums: List[int]) -> List[List[int]]:
+        """回溯的另一种写法"""
+        res = []
+        length = len(nums)
+        def _backtrack(start=0):
+            if start == length:
+                # nums[:] 返回 nums 的一个副本，指向新的引用，这样后续的操作不会影响已经已知解
+                res.append(nums[:])
+            for i in range(start, length):
+                nums[start], nums[i] = nums[i], nums[start]
+                _backtrack(start+1)
+                nums[start], nums[i] = nums[i], nums[start]
+        _backtrack()
+        return res
 ```
 
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@@ -132,4 +151,3 @@ class Solution:
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-