Combination Sum II.txt

problem：
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, � , ak) must be in non-descending order. (ie, a1 ? a2 ? � ? ak).
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] 
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solution：
在candidate numbers中包含重复的元素，因此要使解集合里元素唯一，可能需要对重复元素跳过搜索，但是题目又要求每个元素只能最多被取一次，
因此我们统计某个元素出现最大次数，再进行相应的dfs。开始时还是先进行排序。
//////////////////////////////////////////////////////////////////////////////
void combinationSum_sub(int index, int target, int *times, vector<int> &num, vector<vector<int> > &ret)
{
  if(target < 0)
		return;
	if(index == num.size())
	{
		if(target != 0)
			return;
	}
	if(target == 0)
	{
		vector<int> r;
		for(int i = 0; i < index; i++)  //这里是从0~index-1
			for(int j = 1; j <= times[i]; j++)
				r.push_back(num[i]);
		ret.push_back(r);
		return;
	}
	int maxTimes = 0;
	int indexV = index;
	while(indexV < num.size()  && num[indexV] == num[index])
	{
		maxTimes++;
		indexV++;
	}
	for(int i = 0; i <= maxTimes; i++)
	{
		times[index] = i;
		combinationSum_sub(indexV, target - num[index] * i, times, num, ret);
	}

}

vector<vector<int> > combinationSum2(vector<int> &num, int target)
{
	vector<vector<int> > ret;
	if(num.size() == 0 || target <= 0)
		return ret;
	int *times = new int[num.size()];
	for(int i = 0; i < num.size(); i++)
		times[i] = 0;
	sort(num.begin(), num.end());
	combinationSum_sub(0, target, times, num, ret);
	return ret;
}