Reverse Nodes in Even Length Groups
<p>You are given the <code>head</code> of a linked list.</p>

<p>The nodes in the linked list are <strong>sequentially</strong> assigned to <strong>non-empty</strong> groups whose lengths form the sequence of the natural numbers (<code>1, 2, 3, 4, ...</code>). The <strong>length</strong> of a group is the number of nodes assigned to it. In other words,</p>

<ul>
	<li>The <code>1<sup>st</sup></code> node is assigned to the first group.</li>
	<li>The <code>2<sup>nd</sup></code> and the <code>3<sup>rd</sup></code> nodes are assigned to the second group.</li>
	<li>The <code>4<sup>th</sup></code>, <code>5<sup>th</sup></code>, and <code>6<sup>th</sup></code> nodes are assigned to the third group, and so on.</li>
</ul>

<p>Note that the length of the last group may be less than or equal to <code>1 + the length of the second to last group</code>.</p>

<p><strong>Reverse</strong> the nodes in each group with an <strong>even</strong> length, and return <em>the</em> <code>head</code> <em>of the modified linked list</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2074.Reverse%20Nodes%20in%20Even%20Length%20Groups/images/eg1.png" style="width: 699px; height: 124px;" />
<pre>
<strong>Input:</strong> head = [5,2,6,3,9,1,7,3,8,4]
<strong>Output:</strong> [5,6,2,3,9,1,4,8,3,7]
<strong>Explanation:</strong>
- The length of the first group is 1, which is odd, hence no reversal occurs.
- The length of the second group is 2, which is even, hence the nodes are reversed.
- The length of the third group is 3, which is odd, hence no reversal occurs.
- The length of the last group is 4, which is even, hence the nodes are reversed.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2074.Reverse%20Nodes%20in%20Even%20Length%20Groups/images/eg2.png" style="width: 284px; height: 114px;" />
<pre>
<strong>Input:</strong> head = [1,1,0,6]
<strong>Output:</strong> [1,0,1,6]
<strong>Explanation:</strong>
- The length of the first group is 1. No reversal occurs.
- The length of the second group is 2. The nodes are reversed.
- The length of the last group is 1. No reversal occurs.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2074.Reverse%20Nodes%20in%20Even%20Length%20Groups/images/ex3.png" style="width: 348px; height: 114px;" />
<pre>
<strong>Input:</strong> head = [1,1,0,6,5]
<strong>Output:</strong> [1,0,1,5,6]
<strong>Explanation:</strong>
- The length of the first group is 1. No reversal occurs.
- The length of the second group is 2. The nodes are reversed.
- The length of the last group is 2. The nodes are reversed.
</pre>

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

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>
Linked List
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseEvenLengthGroups(self, head: Optional[ListNode]) -> Optional[ListNode]:
        def reverse(head, l):
            prev, cur, tail = None, head, head
            i = 0
            while cur and i < l:
                t = cur.next
                cur.next = prev
                prev = cur
                cur = t
                i += 1
            tail.next = cur
            return prev

        n = 0
        t = head
        while t:
            t = t.next
            n += 1
        dummy = ListNode(0, head)
        prev = dummy
        l = 1
        while (1 + l) * l // 2 <= n and prev:
            if l % 2 == 0:
                prev.next = reverse(prev.next, l)
            i = 0
            while i < l and prev:
                prev = prev.next
                i += 1
            l += 1
        left = n - l * (l - 1) // 2
        if left > 0 and left % 2 == 0:
            prev.next = reverse(prev.next, left)
        return dummy.next
