# 40. Combination Sum II

## 1.問題

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LO8J0YG1SLWJ3IkQz2Q%2F-LO8K17fC0IxE3UFn8BE%2F%E8%9E%A2%E5%B9%95%E5%BF%AB%E7%85%A7%202018-10-06%20%E4%B8%8B%E5%8D%888.22.00.png?alt=media\&token=04ab8ed0-15cf-47d1-a202-1beed12a322f)

## 2.想法 <a href="#id-2-xiang-fa" id="id-2-xiang-fa"></a>

* 提問
  * 時間複雜度與可用空間是否有限制?&#x20;
  * member是否可以重複使用?
  * 重複的排列是否視為同一個?
* function header, parameter
* test input
* 說明想法
  * 這個問題有點類似39 Combination sum, 但允許member被重複使用
  * 加入限制條件, 當candidates\[i] == candidates\[i - 1]時continue
* 測試計算複雜度

## **3.程式碼** <a href="#id-3-cheng-shi" id="id-3-cheng-shi"></a>

```
class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> record;
        vector<bool> used(candidates.size(), false);
        sort(candidates.begin(), candidates.end());
        getCombinationSum(res, record, candidates, used, target, 0);
        return res;
    }
private:
    void getCombinationSum(vector<vector<int>>& res, vector<int>& record, vector<int>& candidates, vector<bool>& used, int target, int start) {
        if (target < 0) {
            return;
        }
        
        if (target == 0) {
            res.push_back(record);
            return;
        }
        
        for (int i = start; i < candidates.size(); i++) {
            if (i > start && candidates[i] == candidates[i - 1]){
                continue;
            }
            record.push_back(candidates[i]);
            getCombinationSum(res, record, candidates, used, target - candidates[i], i + 1);
            record.pop_back();
        }
    }
};
```