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https://github.com/youngyangyang04/leetcode-master.git
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@ -160,73 +160,34 @@ class Solution {
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```
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## Python
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>直接使用sorted()不符合题意
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>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
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```python
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class Solution:
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def nextPermutation(self, nums: List[int]) -> None:
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"""
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Do not return anything, modify nums in-place instead.
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"""
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for i in range(len(nums)-1, -1, -1):
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for j in range(len(nums)-1, i, -1):
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if nums[j] > nums[i]:
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nums[j], nums[i] = nums[i], nums[j]
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nums[i+1:len(nums)] = sorted(nums[i+1:len(nums)])
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return
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nums.sort()
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```
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>另一种思路
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```python
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class Solution:
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'''
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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。
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故选择另一种算法暂且提供一种python思路
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'''
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def nextPermutation(self, nums: List[int]) -> None:
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"""
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Do not return anything, modify nums in-place instead.
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"""
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length = len(nums)
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for i in range(length-1, 0, -1):
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if nums[i-1] < nums[i]:
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for j in range(length-1, 0, -1):
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if nums[j] > nums[i-1]:
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nums[i-1], nums[j] = nums[j], nums[i-1]
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break
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self.reverse(nums, i, length-1)
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break
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if n == 1:
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# 若正常结束循环,则对原数组直接翻转
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self.reverse(nums, 0, length-1)
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def reverse(self, nums: List[int], low: int, high: int) -> None:
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while low < high:
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nums[low], nums[high] = nums[high], nums[low]
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low += 1
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high -= 1
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```
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>上一版本简化版
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```python
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class Solution(object):
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def nextPermutation(self, nums: List[int]) -> None:
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n = len(nums)
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i = n-2
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while i >= 0 and nums[i] >= nums[i+1]:
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i -= 1
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if i > -1: // i==-1,不存在下一个更大的排列
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j = n-1
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while j >= 0 and nums[j] <= nums[i]:
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j -= 1
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nums[i], nums[j] = nums[j], nums[i]
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for i in range(length - 1, -1, -1):
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for j in range(length - 1, i, -1):
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if nums[j] > nums[i]:
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nums[j], nums[i] = nums[i], nums[j]
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self.reverse(nums, i + 1, length - 1)
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return
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self.reverse(nums, 0, length - 1)
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start, end = i+1, n-1
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while start < end:
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nums[start], nums[end] = nums[end], nums[start]
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start += 1
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end -= 1
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return nums
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def reverse(self, nums: List[int], left: int, right: int) -> None:
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while left < right:
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nums[left], nums[right] = nums[right], nums[left]
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left += 1
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right -= 1
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"""
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265 / 265 个通过测试用例
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状态:通过
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执行用时: 36 ms
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内存消耗: 14.9 MB
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"""
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```
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## Go
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