141. Linked List Cycle
# O(n) time || O(1) space
def has_cycle(self, head: Optional[ListNode]) -> bool:
slow = fast = head
while fast and fast.next:
slow, fast = slow.next, fast.next.next
if slow == fast:
return True
return False
Floyd's Cycle Finding Algorithm by considering two pointers at different speed.
The slow pointer moves one step at a time while the fast pointer moves two steps at a time.