# 回文链表

<p>给你一个单链表的头节点 <code>head</code> ，请你判断该链表是否为回文链表。如果是，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/70b052810fbc7389fc6c77e5f8a7f388.jpeg" style="width: 422px; height: 62px;" />
<pre>
<strong>输入：</strong>head = [1,2,2,1]
<strong>输出：</strong>true
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/2ecd174396cb07fbc62c7c4f0a851623.jpeg" style="width: 182px; height: 62px;" />
<pre>
<strong>输入：</strong>head = [1,2]
<strong>输出：</strong>false
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li>链表中节点数目在范围<code>[1, 10<sup>5</sup>]</code> 内</li>
	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
</ul>

<p>&nbsp;</p>

<p><strong>进阶：</strong>你能否用&nbsp;<code>O(n)</code> 时间复杂度和 <code>O(1)</code> 空间复杂度解决此题？</p>


## template

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

	ListNode(int x) {
		val = x;
	}

}

class Solution {
	public boolean isPalindrome(ListNode head) {

		if (head == null || head.next == null) {
			return true;
		}

		ListNode slow = head;
		ListNode quick = head;

		while (quick != null && quick.next != null) {
			quick = quick.next.next;
			slow = slow.next;
		}
		ListNode pre = null;
		ListNode p = slow;

		while (p != null) {
			ListNode temp = p.next;
			p.next = pre;
			pre = p;
			p = temp;
		}

		while (pre != null) {
			if (pre.val == head.val) {
				pre = pre.next;
				head = head.next;
			} else {
				return false;
			}
		}
		return true;
	}
}
```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```