39. Combination Sum
这道题目应当是一类题目，这类题目有个统一的解法—回溯法(backtracking)

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> combination;
        sort(candidates.begin(),candidates.end());
        combinationSum(candidates, target, res, combination, 0);
        return res;
    }
private:
    void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &res, vector<int> &combination, int begin) 
    {
        if(!target)
        {
            res.push_back(combination);
            return;
        }
        
        for(int i = begin;i < candidates.size() && target >= candidates[i];i++)
        {
            combination.push_back(candidates[i]);
            combinationSum(candidates, target - candidates[i], res, combination, i);
            combination.pop_back();
        }
    }
};

对回溯算法的一个浅显理解有点类似于暴力破解。
回溯法不断的试错，如果当前方法不可行则回退一步或者几步，然后继续尝试不同的答案，直到找到正确答案。