# 90. Subsets II
## 1.問題
## 2.想法
* 提問
* 確認題意: 求數列所有的可能組合, 數列是未經排序, 而且可能有重複數字
* function header, parameter
* test input
* 說明想法
* 做法跟原本的78.subset相同, 但是需要先排序, 而且若num\[i] == num\[i - 1]時則略過
* 測試計算複雜度
## **3.程式碼**
```
class Solution {
public:
vector> subsetsWithDup(vector& nums) {
vector> res;
vector record;
vector used(nums.size(), false);
sort(nums.begin(), nums.end());
for (int i = 1; i <= nums.size(); i++) {
getSubsets(nums, res, i, 0, record, used);
}
res.push_back(vector());
return res;
}
private:
void getSubsets(vector& nums, vector>& res, int n, int start, vector& record, vector& used) {
if (record.size() == n) {
res.push_back(record);
return;
}
for (int i = start; i < nums.size(); i++) {
if (!used[i]) {
if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {
continue;
}
record.push_back(nums[i]);
used[i] = true;
getSubsets(nums, res, n, i + 1, record, used);
record.pop_back();
used[i] = false;
}
}
}
};
```
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