Question:

Given the head of a singly linked list, return the middle node of the linked list.

If there are two middle nodes, return the second middle node.

Example 1:

Input: head = [1,2,3,4,5] Output: [3,4,5] Explanation: The middle node of the list is node 3.

Example 2:

Input: head = [1,2,3,4,5,6] Output: [4,5,6] Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.

Constraints:

The number of nodes in the list is in the range [1, 100]. 1 <= Node.val <= 100

Solution:

• Use the slow and fast pointer technique.
• When the fast pointer reaches the last element or the null pointer, the slow will have reached the middle node.

Code:

class Solution {
    public ListNode middleNode(ListNode head) {

        // You can use two pointers: fast and slow

        ListNode slow = head, fast = head;

        while(fast != null && fast.next != null){

            fast = fast.next.next;

            slow = slow.next;


        }

        return slow;

    }
}

Complexity:

• Time: O(N)
• Space: O(1)