Desctiprion

给定序列，寻找下一个比他大的permutation

Solution

思路：

1. 从后找，到非上升序列时停止
2. (i)/ (i+1)\ ，对于i, 从i+1往后寻找比他大的最小值,并交换（keep order）
3. 最后reverse i+1往后的数组

class Solution:
    def reverse(self,nums,start):
        i = start
        j = len(nums) - 1;
        while i < j:
            tmp = nums[i] 
            nums[i] = nums[j]
            nums [j] = tmp
            i+=1
            j-=1
        
            
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        sub_max = -1
        flag = 0
        i=len(nums)-1
        while i>0:
            if nums[i-1]< nums[i]:  # Not increase from backwards
                break
            i-=1 
        start = i-1
        if i >0:
            # Find the min greater number
            j=i
            while j<len(nums):
                if  nums[start] >= nums[j]:
                    break
                j+=1
            j-=1
            # swap
            tmp = nums[start] 
            nums[start] = nums[j]
            nums [j] = tmp
        self.reverse(nums,start+1)