2095. Delete the Middle Node of a Linked List

1. Description

You are given the head of a linked list. Delete the middle node, and return the head of the modified linked list.
The middle node of a linked list of size n is the ⌊n / 2⌋th node from the start using 0-based indexing, where ⌊x⌋ denotes the largest integer less than or equal to x.

• For n = 1, 2, 3, 4, and 5, the middle nodes are 0, 1, 1, 2, and 2, respectively.

2. Example

Example 1

Input: head = [1,3,4,7,1,2,6]
Output: [1,3,4,1,2,6]
Explanation:
The above figure represents the given linked list. The indices of the nodes are written below.
Since n = 7, node 3 with value 7 is the middle node, which is marked in red.
We return the new list after removing this node.

Example 2

Input: head = [1,2,3,4]
Output: [1,2,4]
Explanation:
The above figure represents the given linked list.
For n = 4, node 2 with value 3 is the middle node, which is marked in red.

Example 3

Input: head = [2,1]
Output: [2]
Explanation:
The above figure represents the given linked list.
For n = 2, node 1 with value 1 is the middle node, which is marked in red.
Node 0 with value 2 is the only node remaining after removing node 1.

3. Constraints

• The number of nodes in the list is in the range [1, 10$^5$].
• 1 <= Node.val <= 10$^5$

4. Solutions

Two Pointers

n is the number of nodes in head
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    ListNode *deleteMiddle(ListNode *head) {
        if (head->next == nullptr) {
            return nullptr;
        }

        ListNode *slow = head;
        ListNode *fast = head->next->next;

        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }

        slow->next = slow->next->next;

        return head;
    }
};