``````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.
``````

``````struct ListNode {
int val;
ListNode *next;
ListNode() : val(0), next(nullptr) {}
ListNode(int x) : val(x), next(nullptr) {}
ListNode(int x, ListNode *next) : val(x), next(next) {}
};

class Solution {
public:
ListNode *middleNode(ListNode *head) {
if (nullptr == head) {
return nullptr;
}
ListNode *slow = head;
ListNode *quick = head;

while (slow != nullptr && quick != nullptr) {
if (quick->next == nullptr || quick->next->next == nullptr) {
if (quick->next == nullptr) {
return slow;
}
return slow->next;
}
quick = quick->next->next;
slow = slow->next;
}
return slow;
}
};
``````