Iterate over the list, re-tying it backwards as you go. If you reach the end, there's no loop. If the order is important, you can re-tie again, still in O(N) time. If you end up at the head, there's a loop.
Fun and really dirty solution! I would never come up with this because I would not dare to modify a passed data structure unless this was the purpose of the procedure.
One could use that answer to gauge the humour of the interviewer :-) For me the expected reaction of a peer to this would be "Cute! Now explain the problems I could encounter when using it." If they can't deal with some fun, well, their loss.
If you can't modify the original list, you can create another as you iterate. But that will require extra O(N) memory, so is worse than the textbook solution with two pointers.
How would you end up there? With a loop, head+1 would point to head after the first pass, so you'll end up at the head. With no loop, head+1 would only be passed once.
