Reverse nodes in even-length linked-list groups

Problem

Given the head of a singly linked list, you will traverse the list in contiguous groups of increasing size: the 1st group has size 1, the 2nd group has size 2, the 3rd group has size 3, and so on.

For each group:

• If the actual number of nodes in that group is even , reverse the nodes in that group.
• If the actual number of nodes in that group is odd , leave the group as-is.

Note: The final group may contain fewer nodes than its intended size; its actual length determines whether it should be reversed.

Return the head of the modified linked list.

Input/Output

• Input: head — head node of a singly linked list.
• Output: head of the updated list after performing the operations above.

Example

List: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9

• Group sizes: 1 | 2 | 3 | 4 ...
• Groups: [1] , [2,3] , [4,5,6] , [7,8,9] (last group has length 3)
• Reverse even-length groups only: [2,3] is even → becomes [3,2] ; others unchanged Result: 1 -> 3 -> 2 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9

Constraints (typical)

• Number of nodes n : 1 <= n <= 10^5
• Node values can be any integers (do not affect the logic).
• Aim for O(n) time and O(1) extra space (excluding recursion stack).