# 876. Middle of the Linked List
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.
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.
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.
Approach 1: fast and slow pointers.
# Code (Python)
class Solution: def middleNode(self, head: ListNode) -> ListNode: if not head: return None fast, slow = head, head while fast and fast.next: fast = fast.next.next slow = slow.next return slow
# Code (C++)