Update 0031.下一个排列.md

完善python代码

Author: casnz1601

problems/0031.下一个排列.md

@@ -160,73 +160,34 @@ class Solution {
 ```
 
 ## Python
->直接使用sorted()不符合题意
+>直接使用sorted()会开辟新的空间并返回一个新的list，故补充一个原地反转函数
 ```python
 class Solution:
     def nextPermutation(self, nums: List[int]) -> None:
         """
         Do not return anything, modify nums in-place instead.
         """
-        for i in range(len(nums)-1, -1, -1):
-            for j in range(len(nums)-1, i, -1):
+        length = len(nums)
+        for i in range(length - 1, -1, -1):
+            for j in range(length - 1, i, -1):
                 if nums[j] > nums[i]:
                     nums[j], nums[i] = nums[i], nums[j]
-                    nums[i+1:len(nums)] = sorted(nums[i+1:len(nums)])
-                    return
-        nums.sort()
-```
->另一种思路
-```python
-class Solution:
-    '''
-    抛砖引玉：因题目要求“必须原地修改，只允许使用额外常数空间”，python内置sorted函数以及数组切片+sort()无法使用。  
-    故选择另一种算法暂且提供一种python思路
-    '''
-    def nextPermutation(self, nums: List[int]) -> None:
-        """
-        Do not return anything, modify nums in-place instead.
-        """
-        length = len(nums)
-        for i in range(length-1, 0, -1):
-            if nums[i-1] < nums[i]:
-                for j in range(length-1, 0, -1):
-                    if nums[j] > nums[i-1]:
-                        nums[i-1], nums[j] = nums[j], nums[i-1]
-                        break
-                self.reverse(nums, i, length-1)
-                break
-            if n == 1:
-                # 若正常结束循环，则对原数组直接翻转
-                self.reverse(nums, 0, length-1)
-
-    def reverse(self, nums: List[int], low: int, high: int) -> None:
-        while low < high:
-            nums[low], nums[high] = nums[high], nums[low]
-            low += 1
-            high -= 1
-```
->上一版本简化版
-```python
-class Solution(object):
-    def nextPermutation(self, nums: List[int]) -> None:
-        n = len(nums)
-        i = n-2
-        while i >= 0 and nums[i] >= nums[i+1]:
-            i -= 1
-
-        if i > -1: // i==-1,不存在下一个更大的排列
-            j = n-1
-            while j >= 0 and nums[j] <= nums[i]:
-                j -= 1
-            nums[i], nums[j] = nums[j], nums[i]
+                    self.reverse(nums, i + 1, length - 1)
+                    return  
+        self.reverse(nums, 0, length - 1)
     
-        start, end = i+1, n-1
-        while start < end:
-            nums[start], nums[end] = nums[end], nums[start]
-            start += 1
-            end -= 1
-        
-        return nums
+    def reverse(self, nums: List[int], left: int, right: int) -> None:
+        while left < right:
+            nums[left], nums[right] = nums[right], nums[left]
+            left += 1
+            right -= 1
+            
+"""
+265 / 265 个通过测试用例
+状态：通过
+执行用时: 36 ms
+内存消耗: 14.9 MB
+"""
 ```
 
 ## Go