[LeetCode] 31. Next Permutation

题目：https://leetcode.com/problems/next-permutation/description/

题目

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

思路

1.这个题目的意思是从这个组合里面，找一个比原组合大一点的数。

分两个部分来看，找到比这个数大，找到比这个数大中最小的。

首先某个较高位需要改变，然后取后面一个比这个数稍大的数与之交换。

2.确定最高位，从后往前找，找到一个前个数比后面一个小的，那么改这前数，是改变数最高位。

3.确定比这个数稍大的数，从后找一个稍大的数。

3.将确定最高位后的数升序排序。

python

class Solution(object):    def nextPermutation(self, nums):        """        :type nums: List[int]        :rtype: void Do not return anything, modify nums in-place instead.        """        if len(nums)==1:            return         subMinindex = -1;        for i in range(len(nums)-1,0,-1):            if nums[i-1]<nums[i]:                subMinindex = i-1                break    print(nums," aaa ",subMinindex)        if subMinindex==-1 :            for i in range(0,len(nums)/2):                nums[i],nums[len(nums)-1-i] = nums[len(nums)-1-i],nums[i]            return         j = subMinindex+2;        while(j < len(nums)):            if nums[subMinindex] >= nums[j]:                break            j+=1;        j = j-1;        nums[subMinindex],nums[j] = nums[j],nums[subMinindex]        nums[subMinindex+1:] = sorted(nums[subMinindex+1:])