[LeetCode] Combination Sum II （递归）

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]

class Solution {
public:
    vector<int> num;
    int target;
    vector<vector<int> > result;
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        this->num = num;
        this->target = target;

        if(num.size()==){
            vector<int> tmp;
            result.push_back(tmp);
            return result;
        }

        sort(this->num.begin(),this->num.end());
        for(int i=;i<num.size();i++){
           vector<int> tmp(,this->num[i]);
           recursion(i+,tmp,this->num[i]);

        }

        return result;
    }
private:
    void recursion(int start,vector<int> tmp,int sum){
        if(sum == target && find(result.begin(),result.end(),tmp)==result.end()){
            result.push_back(tmp);
            return;
        }
        vector<int> tmp0(tmp);
        int sum0 = sum;
        for(int i=start;i<num.size();i++){
            tmp.push_back(num[i]);
            sum += num[i];
            if(sum<target)
               recursion(i+,tmp,sum);
            else if(sum == target){
                if(find(result.begin(),result.end(),tmp)==result.end())
                    result.push_back(tmp);
            }
            else
                break;
            tmp = tmp0;
            sum = sum0;
        }
    }//end func
};