# 子集 II

<p>给你一个整数数组 <code>nums</code> ，其中可能包含重复元素，请你返回该数组所有可能的子集（幂集）。</p><p>解集 <strong>不能</strong> 包含重复的子集。返回的解集中，子集可以按 <strong>任意顺序</strong> 排列。</p><div class="original__bRMd"><div><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>nums = [1,2,2]<strong><br />输出：</strong>[[],[1],[1,2],[1,2,2],[2],[2,2]]</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>nums = [0]<strong><br />输出：</strong>[[],[0]]</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>1 <= nums.length <= 10</code></li>	<li><code>-10 <= nums[i] <= 10</code></li></ul></div></div>

以下程序实现了这一功能，请你填补空白处内容：

```cpp
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
	vector<vector<int>> subsetsWithDup(vector<int> &nums)
	{
		vector<vector<int>> res;
		sort(nums.begin(), nums.end());
		dfs(nums, 0, res);
		return res;
	}
private:
	vector<int> stack;
	void dfs(vector<int> &nums, int start, vector<vector<int>> &res)
	{
		res.push_back(stack);
		int last = INT_MIN;
		for (int i = start; i < nums.size(); i++)
		{
			if (last != nums[i])
			{
				stack.push_back(nums[i]);
				____________________;
				stack.pop_back();
			}
			last = nums[i];
		}
	}
};
```

## template

```cpp
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
	vector<vector<int>> subsetsWithDup(vector<int> &nums)
	{
		vector<vector<int>> res;
		sort(nums.begin(), nums.end());
		dfs(nums, 0, res);
		return res;
	}
private:
	vector<int> stack;
	void dfs(vector<int> &nums, int start, vector<vector<int>> &res)
	{
		res.push_back(stack);
		int last = INT_MIN;
		for (int i = start; i < nums.size(); i++)
		{
			if (last != nums[i])
			{
				stack.push_back(nums[i]);
				dfs(nums, i + 1, res);
				stack.pop_back();
			}
			last = nums[i];
		}
	}
};
```

## 答案

```cpp
dfs(nums, i + 1, res);
```

## 选项

### A

```cpp
dfs(nums, i - 1, res);
```

### B

```cpp
dfs(nums, i, res);
```

### C

```cpp
dfs(nums, 1, res);
```