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

思路：下一个排列，主要依靠字典序来排列的。首先，从最尾端开始往前寻找两个相邻元素，满足num[i]<num[i+1],记录i，找到这样的一组相邻元素，再从最尾端往前查找，找出第一个大于num[i]的元素，索引记为j。将num[i]和num[j]对调，然后将i+1后的所有元素调到排列。记为“下一个”排列组合。

class Solution {

public:

    void nextPermutation(vector<int> &num) {

        int n=num.size();

        if(n<=1)

            return;

        int i=n-2;

        while(num[i]>=num[i+1]&&i>=0)

            i--;

        if(i>=0)

        {

            int j=n-1;

            while(num[i]>=num[j]&&j>=0)

                j--;

            swap(num[i],num[j]);

            reverse(num.begin()+i+1,num.end());

        }

        else

        {

            reverse(num.begin(),num.end());

        }

    }

};