# 90. Subsets II

## 1.問題 

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LNcivxVEXEeloBWtcTW%2F-LNd46ug194fGiwCVYV7%2F%E8%9E%A2%E5%B9%95%E5%BF%AB%E7%85%A7%202018-09-30%20%E4%B8%8B%E5%8D%882.04.37.png?alt=media\&token=4f53386a-70d0-45f7-8ce3-f441b8ba4e22)

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

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

```
class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> record;
        vector<bool> 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<int>());
        return res;
    }
private:
    void getSubsets(vector<int>& nums, vector<vector<int>>& res, int n, int start, vector<int>& record, vector<bool>& 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;
            }
        }
    }
};
```