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

The same repeated number may be chosen from C unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7, 

A solution set is: 

[
  [7],
  [2, 2, 3]
]

class Solution {
public:
	vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
		vector<int>tmp;
		dfs(candidates, target, tmp, 0);
		return s;
	}
private:

	void dfs(vector<int>& candidates, int target, vector<int>&tmp,int cur){
		if (target < 0) return;
		if (target==0){
			s.push_back(tmp);
			return;
		}
		for (int i = cur; i < candidates.size(); i++){
			tmp.push_back(candidates[i]);
			dfs(candidates, target-candidates[i], tmp,i);//i，因為一個數字可以出現多次
			tmp.pop_back();
		}
	}

	vector<vector<int> > s;
};
 

或

class Solution {
public:
	vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
		sort(candidates.begin(), candidates.end());
		vector<int>tmp;
		dfs(candidates, target, tmp, 0);
		return s;
	}
private:

	void dfs(vector<int>& candidates, int target, vector<int>&tmp,int cur){
		if (target < 0) return;
		if (target==0){
			s.push_back(tmp);
			return;
		}
		for (int i = cur; i < candidates.size(); i++){
			if (cur != i&&candidates[i] == candidates[i-1]) continue;
			tmp.push_back(candidates[i]);
			dfs(candidates, target-candidates[i], tmp,i);
			tmp.pop_back();
		}
	}

	vector<vector<int> > s;
};