Given a non-empty, singly linked list with head node head, return a middle node of linked list.

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

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5]) The returned node has value 3.  (The judge's serialization of this node is [3,4,5]). Note that we returned a ListNode object ans, such that: ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6]) Since the list has two middle nodes with values 3 and 4, we return the second one.
Note:The number of nodes in the given list will be between 1 and 100.

Solution:

func middleNode(head *ListNode) *ListNode {
    slow, fast := head, head;
    
    for fast != nil && fast.Next != nil {
        slow = slow.Next
        fast = fast.Next.Next
    }
    return slow;
}