Linked List Cycle

Advance slow by one node and fast by two. In an acyclic list, fast reaches null. In a cycle, the speed difference of one eventually makes fast catch slow, so equality proves a cycle. The test is O(n) time and O(1) space. To find the entry after a meeting, reset one pointer to head and move both one step at a time; their next meeting is the entry. Guard both fast and fast.next before advancing. In an interview, make the invariant explicit before writing code, trace a boundary example that breaks a tempting shortcut, state what each stored value represents, and explain why every update preserves correctness. This verification step also surfaces empty-input, duplicate-value, and numeric-limit cases.