LeetCode Next Permutation

题意

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.
求出下一个排列

解法

从后往前看，如果一个数后面的所有数字都比它小，那么后面的排列就已经是最终形式，不用再调整了。相反，如果后面有数比它大，那么就交换他们的位置。

class Solution
{
public:
    void nextPermutation(vector<int>& nums)
    {
	    bool	flag = false;

	    for(int i = nums.size() - 1;i >= 0;i --)
	    {
		    for(int j = nums.size() - 1;j > i;j --)
			    if(nums[j] > nums[i])
			    {
				    swap(nums[i],nums[j]);
				    flag = true;
				    sort(nums.begin() + i + 1,nums.end());
				    break;
			    }
		    if(flag)
			    break;
	    }
	    if(!flag)
		    sort(nums.begin(),nums.end());
    }
};