Unlink the Later Middle Node

### Problem Given the head of a singly linked list with `n` nodes, remove the node at zero-based position `floor(n / 2)` and return the head of the remaining list. For even `n`, this means the later of the two middle nodes. A one-node list becomes empty. Keep the other nodes in their original order. ```cpp class Solution { public: ListNode* deleteMiddle(ListNode* head); }; ``` ### Examples **Example 1:** ```text List values: [8,6,4,2,1] Output list values: [8,6,2,1] Explanation: The removed position is floor(5/2)=2, holding 4. ``` **Example 2:** ```text List values: [7] Output: nullptr Explanation: Position zero is the middle and is removed. ``` ### Constraints - The list has 1 to `10^5` nodes. - Node values lie in `[1,10^5]`. ### Clarifying Questions Delete the node from the reachable chain by changing links; do not replace values to imitate deletion. Memory ownership/deallocation is not specified by the source. ```hint Inspect the structure Where is a one-step pointer when a two-step pointer reaches the end? ``` ### Reasoning checkpoints - Remove exactly the specified middle position for both odd and even list lengths. - Preserve the remaining nodes' identities, values and relative order. - Handle a singleton without dereferencing a missing predecessor or successor. ### Optional extensions How would you adapt the solution if the earlier middle were requested for an even-length list?