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# 40. Combination Sum II
## 1.問題
## 2.想法
* 提問
* 時間複雜度與可用空間是否有限制?
* 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**
```
class Solution {
public:
vector> combinationSum2(vector& candidates, int target) {
vector> res;
vector record;
vector used(candidates.size(), false);
sort(candidates.begin(), candidates.end());
getCombinationSum(res, record, candidates, used, target, 0);
return res;
}
private:
void getCombinationSum(vector>& res, vector& record, vector& candidates, vector& 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();
}
}
};
```
---
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