# 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>

## template

```java
class Solution {
	public List<List<Integer>> combinationSum(int[] candiates, int target) {

		List<List<Integer>> resultList = new ArrayList<>();

		List<Integer> result = new ArrayList<>();

		Arrays.sort(candiates);

		dfs(candiates, resultList, result, 0, target);
		return resultList;

	}

	private void dfs(int[] candiates, List<List<Integer>> resultList, List<Integer> result, int start, int target) {

		if (target < 0) {
			return;
		}

		else if (target == 0) {

			resultList.add(new ArrayList<>(result));
		} else {
			for (int i = start; i < candiates.length; i++) {
				result.add(candiates[i]);

				dfs(candiates, resultList, result, i, target - candiates[i]);

				result.remove(result.size() - 1);
			}
		}
	}
}

```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```