# 39 - Combination Sum

解法一 - Backtracking

雖然同樣一個 element 可以重複用無限次，但也要小心避免重複的答案被加進 sol，所以我們需要多傳一個 begin 進去以避免從之前已經找過的開頭重複找。

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> sol;
        vector<int> state;

        sort(candidates.begin(), candidates.end());
        backtrack(candidates, target, state, sol, 0);

        return sol;
    }

private:
    void backtrack(vector<int>& candidates, int target, vector<int>& state, vector<vector<int>>& sol, int begin) {
        if(target == 0)
            sol.push_back(state);
        else if (target < 0)
            return;
        else
        {
            for(int i=begin; i<candidates.size(); i++)
            {
                state.push_back(candidates[i]);
                backtrack(candidates, target-candidates[i], state, sol, i);
                state.pop_back();
            }
        }
    }
};