# 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

```python

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        current = head
        nextNode = None
        h = head
        while current is not None and current.next is not None:
            nextNode = current.next
            if nextNode.next is not None:
                tmpNode = current.next
                current.next = nextNode.next
                tmpNode.next = h

            else:
                current.next = None
                nextNode.next = h
            h = nextNode

        return h

```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```