LeetCode_Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a₁, a₂, � , a_k) must be in non-descending order. (ie, a₁ ? a₂ ? � ? a_k).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

vector > combinationSum2(vector &num, int target) 
{
    vector > ans;
	int size = num.size();
	if (size == 0)
	{
		return ans;
	}

    sort(num.begin(), num.end());
	int remain = 0;
	for (int i = 0; i < size; ++i)
	{
		remain += num[i];
	}
	vector vec;
	DFS(ans, num, 0, target, vec, size, 0, remain);
	sort(ans.begin(), ans.end());
	int total = ans.size();
	for (int i = 1; i < total; ++i)
	{
		int t1 = ans[i-1].size();
		int t2 = ans[i].size();
		bool isSame = true;
		if (t1 != t2)
		{
			continue;
		}
		for (int j = 0; j < t1; ++j)
		{
			if (ans[i-1][j] != ans[i][j])
			{
				isSame = false;
				break;
			}
		}
		if (isSame)
		{
			ans.erase(ans.begin() + i);
			--total;
			--i;
		}
			
	}
	return ans;
}

void DFS(vector >& ans, vector &num, int sum, 
	int target, vector &vec, int size, int from, int remain)
{
	for (int i = from; i < size; ++i)
	{			
		if (sum + num[i] <= target)
		{			
			vec.push_back(num[i]);
			if (sum + num[i] == target)
			{
				ans.push_back(vec);
				vec.erase(vec.end() - 1);
				break;
			}
			DFS(ans, num, sum + num[i], target, vec, size, i + 1, remain - num[i]);
			vec.erase(vec.end() - 1);
		}

	}
}