since it's java, you have access to a Set. add each node to the set if it is not present, O(1). A bit more aggressive, check System.Runtime.heapSize() calculate

since it's java, you have access to a Set. add each node to the set if it is not present, O(1). A bit more aggressive, check System.Runtime.heapSize() calculate how many nodes can fit in the heap, and subtract 1 for each element visited. Still O(1) With lower level access, recognize that pointers are always powers of 2, so go ahead and tag that pointer (p = p | 1, if *p & 1, it's a cycle.) you'll build up a stack, and have to fix the tagged pointers, but also O(1). I dunno. the double advance is a cute trick, but if the fast moving pointer doesn't do the check, you can easily spin into an infinite loop. It's not a _hard_ problem. You just have to keep your wits about you. Lots of ways to solve it.