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 ≤ … ≤ 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] 

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        //回溯法
        sort(num.begin(),num.end());  
        vector< vector<int> > res;
        vector <int> path;
        solve(num,0,0,target,res,path);
        return res;
    }
    
     void solve( vector<int> candidates,int index,int sum,int target, vector< vector<int> > &res, vector<int> &path)
    {
        if(sum>target)
            return ;
        if(sum==target)
        {
            res.push_back(path);
            return ;
        }
        for(int i=index;i<candidates.size();i++)
        {
            sum+=candidates[i];
            path.push_back(candidates[i]);
            solve(candidates,i+1,sum,target,res,path);
            path.pop_back();
            sum-=candidates[i];
            while(i<candidates.size()-1 && candidates[i]==candidates[i+1])i++;  
        }
    }
       
};