思路：兩種方法都是利用遞歸回溯，第二方法在第一種方法的基礎上對原始資料先進行排序，這樣可以剪枝，加快計算速度。第一種方法在LeetCode上測試執行的時間是24ms，第二種方法執行時間為16ms。
方法一：

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

class Solution {
private:
    //res儲存滿足題意的結果
    vector<vector<int>> res;
    //儲存當前的一種方案
 

    vector<int> method;
public:
    void getCombinationSum(int start, int sum, int target, vector<int> &candidates)
    {
            //因為canditates經過排序，所以如果sum>target
            if (sum > target)
            {
                return;
            }
            //如果滿足題意則儲存結果
            else
 
 if (sum == target)
            {
                res.push_back(method);
                return;
            } 
            else
            {
                for(int i=start;i<candidates.size();++i)
                {
                    //芳第i件物品
                    method.push_back(candidates[i]);
                    getCombinationSum(i, sum+candidates[i], target, candidates);
                    //回溯，去除第i件物品，然後繼續放第i+1個數
 

                    method.pop_back();
                }
            }
    }

    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        int sum = 0;
        getCombinationSum(0, 0, target, candidates);
        return res;
    }
};

int main()
{
    Solution s;
    vector<int> canditates;
    canditates.push_back(2);
    canditates.push_back(7);
    canditates.push_back(3);
    canditates.push_back(6);

    vector<vector<int>> res = s.combinationSum(canditates, 7);
    //列印結果
    for (vector<vector<int>>::iterator ita=res.begin();ita!=res.end();++ita)
    {
        for (vector<int>::iterator itb=ita->begin();itb!=ita->end();++itb)
        {
            cout<<*itb<<'\t';
        }
        cout<<endl;
    }
    cout<<endl;
    return 0;
}

方法二：

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

class Solution {
private:
    //res儲存滿足題意的結果
    vector<vector<int>> res;
    //儲存當前的一種方案
    vector<int> method;
public:
    void getCombinationSum(int start, int sum, int target, vector<int> &candidates)
    {
        for(int i=start;i<candidates.size();++i)
        {
            //放第i個數,如果第i個數放進去之後超過了target，則剪枝（直接返回）,因為candidates經過排序，換成後面的數放進去的話肯定也會超過target
            if (sum+candidates[i] > target)
            {
                return;
            }
            //如果滿足則儲存結果
            else if (sum+candidates[i] == target)
            {
                method.push_back(candidates[i]);
                res.push_back(method);
                method.pop_back();
                return;
            }
            //如果放第i個數後不超過target，則將第i個數放入
            else
            {
                method.push_back(candidates[i]);
                getCombinationSum(i, sum+candidates[i], target, candidates);
                //回溯，去除第i個數，然後繼續放第i+1個數
                method.pop_back();
            }                   
        }
    }

    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        int sum = 0;
        //排序
        sort(candidates.begin(), candidates.end());
        getCombinationSum(0, 0, target, candidates);
        return res;
    }
};

int main()
{
    Solution s;
    vector<int> canditates;
    canditates.push_back(2);
    canditates.push_back(7);
    canditates.push_back(3);
    canditates.push_back(6);

    vector<vector<int>> res = s.combinationSum(canditates, 7);
    for (vector<vector<int>>::iterator ita=res.begin();ita!=res.end();++ita)
    {
        for (vector<int>::iterator itb=ita->begin();itb!=ita->end();++itb)
        {
            cout<<*itb<<'\t';
        }
        cout<<endl;
    }
    cout<<endl;
    return 0;
}