python-leetcode-下一个排列

31. 下一个排列 - 力扣（LeetCode）

class Solution:def nextPermutation(self, nums: List[int]) -> None:"""Do not return anything, modify nums in-place instead."""# Step 1: Find the first decreasing element from the righti = len(nums) - 2while i >= 0 and nums[i] >= nums[i + 1]:i -= 1if i >= 0:# Step 2: Find the element that is larger than nums[i] from the rightj = len(nums) - 1while nums[j] <= nums[i]:j -= 1# Step 3: Swap nums[i] and nums[j]nums[i], nums[j] = nums[j], nums[i]# Step 4: Reverse the part of the list from i + 1 to endnums[i + 1:] = reversed(nums[i + 1:])