solution.md

# 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

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

class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: None Do not return anything, modify head in-place instead.
        """

        if not head:
            return

        stack = []
        s = head

        while s.next:
            stack.append(s.next)
            s = s.next

        s = head

        n = 0
        while stack:
            if n % 2 == 0:
                one = stack.pop()
            else:
                one = stack.pop(0)
            one.next = None
            s.next = one
            s = s.next
            n += 1


```

## 答案

```python

```

## 选项

### A

```python

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### B

```python

```

### C

```python

```