递归（2）

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.
• 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]]

解析：此题主要运用递归来解决问题，以上周一题极为相似，因此一同做了。

程序：

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) 
    {
        vector<vector<int>> res;
        sort(num.begin(),num.end());
        vector<int> local;
        findCombination(res, 0, target, local, num);
        return res;
    }
    void findCombination(vector<vector<int>>& res, const int order, const int target, vector<int>& local, const vector<int>& num)
    {
        if(target==0)
        {
            res.push_back(local);
            return;
        }
        else
        {
            for(int i = order;i<num.size();i++) // iterative component
            {
                if(num[i]>target) return;
                if(i&&num[i]==num[i-1]&&i>order) continue; // check duplicate combination
                local.push_back(num[i]),
                findCombination(res,i+1,target-num[i],local,num); // recursive componenet
                local.pop_back();
            }
        }
    }
};