Exercise 1 solution

    class Node:
        def __init__(self, value, next=None):
            self.value = value
            self.next = next

    def build(values):
        head = None
        for v in reversed(values):
            head = Node(v, head)
        return head

    def middle(head):
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow.value if slow else None

    print(middle(build([1, 2, 3, 4, 5])), middle(build([1, 2, 3, 4])),
          middle(build([7])), middle(None))

Output:
    3 3 7 None

The fast pointer moves two nodes for every one the slow pointer moves, so
when fast reaches the end, slow is halfway. One pass, no length needed.
For an even number of nodes this returns the second of the two middle
nodes (3 in 1-2-3-4). The "while fast and fast.next" test covers the empty
list, a single node, and both parities without special cases.