Your question is Find Middle of Linked List. Start with the requirements on the right.
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An Edwards Lifesciences monitoring component stores a sequence of numeric readings in a singly linked list. Find and return the middle reading using a single traversal and constant extra space.
Represent each node as a dictionary with a value field and a next field. The next field contains another node dictionary or None. If the list has an even number of nodes, return the second of the two middle values. Return None when the list is empty.
Implement middle_number(head), where head is either None or the first node in a singly linked list. Each node has an integer value and a next pointer. Return the integer stored in the middle node, or None for an empty list.
Use the slow and fast pointer technique. The slow pointer advances one node at a time, while the fast pointer advances two nodes at a time.
def middle_number(head):