Reorder List
<p>You are given the head of a singly linked-list. The list can be represented as:</p>

<pre>
L<sub>0</sub> &rarr; L<sub>1</sub> &rarr; &hellip; &rarr; L<sub>n - 1</sub> &rarr; L<sub>n</sub>
</pre>

<p><em>Reorder the list to be on the following form:</em></p>

<pre>
L<sub>0</sub> &rarr; L<sub>n</sub> &rarr; L<sub>1</sub> &rarr; L<sub>n - 1</sub> &rarr; L<sub>2</sub> &rarr; L<sub>n - 2</sub> &rarr; &hellip;
</pre>

<p>You may not modify the values in the list&#39;s nodes. Only nodes themselves may be changed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0143.Reorder%20List/images/reorder1linked-list.jpg" style="width: 422px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4]
<strong>Output:</strong> [1,4,2,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0143.Reorder%20List/images/reorder2-linked-list.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5]
<strong>Output:</strong> [1,5,2,4,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 5 * 10<sup>4</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 1000</code></li>
</ul>
Stack; Recursion; Linked List; Two Pointers
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: Optional[ListNode]) -> None:
        fast = slow = head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next

        cur = slow.next
        slow.next = None

        pre = None
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        cur = head

        while pre:
            t = pre.next
            pre.next = cur.next
            cur.next = pre
            cur, pre = pre.next, t
