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

刚开始没看懂，后面查看网上资料才搞懂，就是转化为按照字典顺序排列的后面的一个排列

如:15467543应该为15473456

class Solution {
public:
    void nextPermutation(vector<int> &num) {
       int i, j;
	i = num.size() - 2;
	j = num.size() - 1;
	//从后往前找，先找到不满足非降序的第一个数；
	while(i >= 0 && num[i] >= num[i + 1])
	{
		i--;
	}
	//从后往前找，找到大于上面找到的非降序的第一个数
	while(j >= 0 && num[j] <= num[i])
	{
		j--;
	}
	if(i < j)
	{
		int temp;
		temp = num[j];
		num[j] = num[i];
		num[i] = temp;
		//swap(num[i], num[j]);
		sort(num.begin() + i + 1, num.end());
	}
	else
	{
		reverse(num.begin(), num.end());
	}
    }
};