First, eliminate the pathological case of a 3-element list with a cycle from element 2 to 1. Keep two 
pointers. One at element 1 and the next at element 3. See if they are equal; if not, move P1 by one and 
P2 by two. Check and continue. If P1 or P2 is null, there is no loop. If there is one, it will definitely be 
detected. One pointer will eventually catch up with the other (i.e., have the same value), though it 
might take several traversals of the cycle to do it.