# 反转链表

给你单链表的头节点 <code>head</code> ，请你反转链表，并返回反转后的链表。
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<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/1479f99ec2d8583971cc3dfb0c59e0cb.jpeg" style="width: 542px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2,3,4,5]
<strong>输出：</strong>[5,4,3,2,1]
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/1b87f6d74f7125930f57958e210e996b.jpeg" style="width: 182px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2]
<strong>输出：</strong>[2,1]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>head = []
<strong>输出：</strong>[]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>链表中节点的数目范围是 <code>[0, 5000]</code></li>
	<li><code>-5000 <= Node.val <= 5000</code></li>
</ul>

<p> </p>

<p><strong>进阶：</strong>链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？</p>
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## template

```java
public class ListNode {
	int val;
	ListNode next;

	ListNode(int x) {
		val = x;
	}
}

class Solution {
	public ListNode reverseList(ListNode head) {
		ListNode pre = null;
		while (head != null) {

			ListNode tmp = head.next;

			head.next = pre;

			pre = head;

			head = tmp;
		}
		return pre;
	}
}
```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```