leetcode-31-下一个排列

题目描述：

方法一：O(n) O(1)

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        firstIndex = -1
        n = len(nums)
        def reverse(nums,i,j):
            while i<j:
                nums[i],nums[j] = nums[j],nums[i]
                i += 1
                j -= 1
            return nums
        for i in range(n-2, -1, -1):
            if nums[i] < nums[i+1]:
                firstIndex = i
                break
        if firstIndex == -1:
            reverse(nums,0,n-1)
            return
        secondIndex = -1
        for i in range(n-1, firstIndex, -1): 
            if nums[i] > nums[firstIndex]: 
                secondIndex = i 
                break
        nums[firstIndex],nums[secondIndex] = nums[secondIndex], nums[firstIndex] 
        reverse(nums, firstIndex+1, n-1)