# 重排链表
给定一个单链表 L 的头节点 head ，单链表 L 表示为：
L0 → L1 → … → Ln-1 → Ln
请将其重新排列后变为：
L0 → Ln → L1 → Ln-1 → L2 → Ln-2 → …
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。
示例 1:
输入: head = [1,2,3,4]
输出: [1,4,2,3]
示例 2:
输入: head = [1,2,3,4,5]
输出: [1,5,2,4,3]
提示：
- 链表的长度范围为
[1, 5 * 104]
1 <= node.val <= 1000
## 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
```