Combination Sum

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.

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 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

思路：我们通过backtracking来记录所有的值。这道题里用一个函数assist来协助完成。

对于这一串数，我们先将其排序，然后枚举每一个小于当前target的位置。假如现在枚举到了第i个数，则递归寻找从第i位开始所有累加和为target - candidates[i]的可能。

public:
    void assist(vector<int>& candidates, vector<vector<int> >& res, vector<int>& cur, int target, int st)
    {
        if (target == 0)
        {
            res.push_back(cur);
            return;
        }
        for (int i = st, n = candidates.size(); i < n && candidates[i] <= target; i++)
        {
            cur.push_back(candidates[i]);
            assist(candidates, res, cur, target - candidates[i], i);
            cur.pop_back();
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int> > res;
        vector<int> cur;
        sort(candidates.begin(), candidates.end(), less<int>());
        assist(candidates, res, cur, target, 0);
        return res;
    }
};