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 right
i = len(nums) - 2
while i >= 0 and nums[i] >= nums[i + 1]:
i -= 1
if i >= 0:
# Step 2: Find the element that is larger than nums[i] from the right
j = len(nums) - 1
while 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 end
nums[i + 1:] = reversed(nums[i + 1:])
