31. Next Permutation leetcode Python 2016 new Season

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3

1,1,5 → 1,5,1




class Solution(object):
    def reverse(self, nums, start_index):
        left_index = start_index
        right_index = len(nums) - 1
        while left_index < right_index:
            nums[left_index], nums[right_index] = nums[right_index], nums[left_index]
            left_index += 1
            right_index -= 1

    def nextPermutation(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        if not nums:
            return
        first_mark = len(nums) - 2
        # 2 3 6 5 4 1
        # back trace find first number that does not increase in order => 3
        while first_mark >= 0 and nums[first_mark] >= nums[first_mark + 1]:
            first_mark -= 1
        if first_mark >= 0:
            second_mark = first_mark + 1
        # find last number greater than 3 => 4 
            while second_mark < len(nums) and nums[second_mark] > nums[first_mark]:
                second_mark += 1
            second_mark -= 1
        # swop the two numbers
            nums[first_mark], nums[second_mark] = nums[second_mark], nums[first_mark]
        # reverse all the numbers after the second mark
        self.reverse(nums, first_mark + 1)