题目：

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] 

思路：

常规递归问题。

代码：

class Solution {
public:
    vector<vector<int> > result;
    vector<int> v;
    vector<int> candidates;
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        this->candidates = num;
        sort(this->candidates.begin(),this->candidates.end());
        getCombination(0,0,target);
        return result;
    }
    
    void getCombination(int k, int sum, int target)
    {
        if(sum==target)
        {
            for(int k=0;k<result.size();k++)
            {
                if(result[k]==v)
                {
                    return;
                }
            }
            result.push_back(v);
        }
        else if(sum>target)
        {
            return;
        }
        else if(k>=candidates.size())
        {
            return;
        }
        else
        {
            getCombination(k+1, sum, target);
            v.push_back(candidates[k]);
            getCombination(k+1, sum+candidates[k], target);
            v.pop_back();
        }
    }
};