【leetcode】Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

说明：

 bool nextPermutetion(vector<int> &num)
{
	int i = num.size() - 1;
	while (i >= 1)
	{
		if(num[i] > num[i - 1])
		{
			--i;
			int ii = num.size() - 1;
			while (ii > i && num[ii] <= num[i]) --ii;
			if(ii > i)
			{
				swap(num[i], num[ii]);
				reverse(num.begin() + i + 1, num.end());

				return true;
			}

		}
		else
			--i;
	}
	if(i == 0)
		return false;
	return true;
}

vector<vector<int> > permuteUnique(vector<int> &num) {
	vector<vector<int> > re;
	if(num.size() == 0)
		return re;
	sort(num.begin(), num.end());
	re.push_back(num);
	while (nextPermutetion(num))
	{
		re.push_back(num);
	}
	return re;
}

vector<vector<int> > permuteUnique(vector<int> &num) {
	vector<vector<int> > re;
	if(num.size() == 0)
		return re;
	sort(num.begin(), num.end());
	re.push_back(num);
	int i = num.size() - 1;
	
	while(1){
	    int ii = i;
	    --i;
	    if(i < 0)
	    return re;
		if(num[ii] > num[i])
		{
			ii = num.size() - 1;
			while (ii > i && num[ii] <= num[i]) --ii;
			if(ii > i)
			{
				swap(num[i], num[ii]);
				reverse(num.begin() + i + 1, num.end());
				re.push_back(num);
				i = num.size() - 1;
			}
		}
	}//end while 1
	return re;
}