# 重排链表

<p>给定一个单链表 <code>L</code><em> </em>的头节点 <code>head</code> ，单链表 <code>L</code> 表示为：</p>

<p><code> L<sub>0 </sub>→ L<sub>1 </sub>→ … → L<sub>n-1 </sub>→ L<sub>n </sub></code><br />
请将其重新排列后变为：</p>

<p><code>L<sub>0 </sub>→ L<sub>n </sub>→ L<sub>1 </sub>→ L<sub>n-1 </sub>→ L<sub>2 </sub>→ L<sub>n-2 </sub>→ …</code></p>

<p>不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。</p>

<p> </p>

<p><strong>示例 1:</strong></p>

<p><img alt="" src="https://pic.leetcode-cn.com/1626420311-PkUiGI-image.png" style="width: 240px; " /></p>

<pre>
<strong>输入: </strong>head = [1,2,3,4]
<strong>输出: </strong>[1,4,2,3]</pre>

<p><strong>示例 2:</strong></p>

<p><img alt="" src="https://pic.leetcode-cn.com/1626420320-YUiulT-image.png" style="width: 320px; " /></p>

<pre>
<strong>输入: </strong>head = [1,2,3,4,5]
<strong>输出: </strong>[1,5,2,4,3]</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>链表的长度范围为 <code>[1, 5 * 10<sup>4</sup>]</code></li>
	<li><code>1 <= node.val <= 1000</code></li>
</ul>


## template

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

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

class Solution {
	public void reorderList(ListNode head) {
		if (head == null) {
			return;
		}
		ListNode fast = head, slow = head;
		while (fast != null && fast.next != null) {
			fast = fast.next.next;
			slow = slow.next;
		}

		ListNode cur = slow.next, pre = null, next = null;
		slow.next = null;
		while (cur != null) {
			next = cur.next;
			cur.next = pre;
			pre = cur;
			cur = next;
		}

		ListNode p1 = head, p2 = pre;
		while (p1 != null && p2 != null) {
			next = p2.next;
			p2.next = p1.next;
			p1.next = p2;
			p1 = p2.next;
			p2 = next;
		}
	}
}

```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```