# 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

```python
class Solution(object):
	def combinationSum(self, candidates, target):
		candidates.sort()
		dp = [[] for _ in range(target + 1)]
		dp[0].append([])
		for i in range(1, target + 1):
			for j in range(len(candidates)):
				if candidates[j] > i:
					break
				for k in range(len(dp[i - candidates[j]])):
					temp = dp[i - candidates[j]][k][:]
					if len(temp) > 0 and temp[-1] > candidates[j]:
						continue
					temp.append(candidates[j])
					dp[i].append(temp)
		return dp[target]
if __name__ == '__main__':
	s = Solution()
	print (s.combinationSum([8,7,4,3], 11))
```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```