Detect Cycle in Linked List

Use Floyd's tortoise-and-hare algorithm. Start slow and fast at the head; move slow one step and fast two steps while fast and fast.next exist. If fast reaches null, there is no cycle. If they meet, a cycle exists because, once both are inside it, their relative speed is one node per iteration. To find the entry, reset one pointer to head and keep the other at the meeting point, then move both one step; their next meeting is the entry. The proof uses distances: if the non-cycle prefix has length mu and cycle length lambda, the first meeting has travelled a multiple relationship that makes both pointers mu steps from the entry after reset. A hash set of visited node identities is simpler but costs O(n) memory. Test an empty list, one node with and without self-loop, a cycle beginning at head, and a long acyclic prefix. Runtime is O(n) and Floyd uses O(1) extra space.