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.

For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

1.​​dfs(candidates,target,i,tmp,sum);​​​ 下次迭代还是从i开始，这样值可以重复
2.怎么避免得到的向量重复，有两种方法：
（1）

if(i>0 && candidates[i] == candidates[i-1])//deal with dupicate
                 continue;

（2）

if(!res.contains(item)) 
                res.add(new ArrayList<Integer>(item));
class Solution {
public:
    /**
     * @param candidates: A list of integers
     * @param target:An integer
     * @return: A list of lists of integers
     */
    vector<vector<int> >res;

    void dfs(vector<int> &candidates, int target,int cur,vector<int> &tmp,int sum){

        if(sum>target){
            return;
        }

        if(sum==target){
            res.push_back(tmp);
            return;
        }

        for(int i=cur;i<candidates.size();i++){
            if(i>0 && candidates[i]==candidates[i-1]){
                continue;
            }

            tmp.push_back(candidates[i]);
            sum+=candidates[i];

            dfs(candidates,target,i,tmp,sum);//i

            sum-=candidates[i];
            tmp.pop_back();

        }
    }

    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // write your code here

        vector<int> tmp;

        sort(candidates.begin(),candidates.end());

        dfs(candidates,target,0,tmp,0);

        return res;
    }
};