# 组合总和

<p>给定一个<strong>无重复元素</strong>的数组&nbsp;<code>candidates</code>&nbsp;和一个目标数&nbsp;<code>target</code>&nbsp;，找出&nbsp;<code>candidates</code>&nbsp;中所有可以使数字和为&nbsp;<code>target</code>&nbsp;的组合。
</p>
<p><code>candidates</code>&nbsp;中的数字可以无限制重复被选取。</p>
<p><strong>说明：</strong></p>
<ul>
    <li>所有数字（包括&nbsp;<code>target</code>）都是正整数。</li>
    <li>解集不能包含重复的组合。&nbsp;</li>
</ul>
<p><strong>示例&nbsp;1：</strong></p>
<pre><strong>输入：</strong>candidates = [2,3,6,7], target = 7,<strong><br />输出：</strong>[[7],[2,2,3]]</pre>
<p><strong>示例&nbsp;2：</strong></p>
<pre><strong>输入：</strong>candidates = [2,3,5], target = 8,<strong><br />输出：</strong>[[2,2,2,2],[2,3,3],[3,5]]</pre>
<p>&nbsp;</p>
<p><strong>提示：</strong></p>
<ul>
    <li><code>1 &lt;= candidates.length &lt;= 30</code></li>
    <li><code>1 &lt;= candidates[i] &lt;= 200</code></li>
    <li><code>candidate</code> 中的每个元素都是独一无二的。</li>
    <li><code>1 &lt;= target &lt;= 500</code></li>
</ul>

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>> combinationSum(vector<int> &candidates, int target)
	{
		vector<vector<int>> res;
		dfs(candidates, 0, target, res);
		return res;
	}
private:
	vector<int> stack;
	void dfs(vector<int> &candidates, int start, int target, vector<vector<int>> &res)
	{
		if (target < 0)
		{
			return;
		}
		else if (target == 0)
		{
			res.push_back(stack);
		}
		else
		{
			for (int i = start; i < candidates.size(); i++)
			{
				stack.push_back(candidates[i]);
				_____________________________;
				stack.pop_back();
			}
		}
	}
};
```

## template

```cpp
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
	vector<vector<int>> combinationSum(vector<int> &candidates, int target)
	{
		vector<vector<int>> res;
		dfs(candidates, 0, target, res);
		return res;
	}
private:
	vector<int> stack;
	void dfs(vector<int> &candidates, int start, int target, vector<vector<int>> &res)
	{
		if (target < 0)
		{
			return;
		}
		else if (target == 0)
		{
			res.push_back(stack);
		}
		else
		{
			for (int i = start; i < candidates.size(); i++)
			{
				stack.push_back(candidates[i]);
				dfs(candidates, i, target - candidates[i], res);
				stack.pop_back();
			}
		}
	}
};
```

## 答案

```cpp
dfs(candidates, i, target - candidates[i], res);
```

## 选项

### A

```cpp
dfs(candidates, i, target - i, res);
```

### B

```cpp
dfs(candidates, i, candidates[i], res);
```

### C

```cpp
dfs(candidates, i, target + candidates[i], res);
```