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).