Next Permutation

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

方法：
参考网站：
https://discuss.leetcode.com/topic/15216/a-simple-algorithm-from-wikipedia-with-c-implementation-can-be-used-in-permutations-and-permutations-ii。

从数组的右侧进行遍历，找出第一个nums[i] < nums[i+1]的数，将该位置标记为pos1。然后在大于pos1的位置上找出最后一个大于nums[pos1]的位置，记为pos2。将这两个位置上的数字进行调换，然后将大于pos1位置的数字进行逆置。

class Solution {public:    void nextPermutation(vector<int>& nums) {        int last = nums.size() - 2;        while(last >=0){            if(nums[last]>=nums[last+1])                --last;            else{                break;            }        }        if(last < 0){            sort(nums.begin(),nums.end());            return;        }        else{            int index = -1;            for(int j = nums.size() -1 ; j >=last; --j){                if(nums[j]>nums[last]){                    swap(nums[j],nums[last]);                    break;                }            }        }        reverse(nums.begin()+last+1,nums.end());        return ;    }};