How do you determine if your singly-linked list has a cycle? In 1980, Brent invented an algorithm that not only worked in linear time, but required less stepping than Floyd's Tortoise and the Hare algorithm (however it is slightly more complex). Although stepping through a 'regular' linked list is computationally easy, these algorithms are also used for factorization and pseudorandom number generators, linked lists are implicit and finding the next member is computationally difficult.

Brent's algorithm features a moving rabbit and a stationary, then teleporting, turtle. Both turtle and rabbit start at the top of the list. The rabbit takes one step per iteration. If it is then at the same position as the stationary turtle, there is obviously a loop. If it reaches the end of the list, there is no loop.

Of course, this by itself will take infinite time if there is a loop. So every once in a while, we teleport the turtle to the rabbit's position, and let the rabbit continue moving. We start out waiting just 2 steps before teleportation, and we double that each time we move the turtle.

Why move the turtle at all? Well, the loop might not include the entire list; if a rabbit gets stuck in a loop further down, without the turtle, it will go forever. Why take twice as long each time? Eventually, the length of time between teleportations will become longer than the size of the loop, and the turtle will be there waiting for the rabbit when it gets back.

Note that like Floyd's Tortoise and Hare algorithm, this one runs in O(N). However you're doing less stepping than with Floyd's (in fact the upper bound for steps is the number you would do with Floyd's algorithm). According to Brent's research, his algorithm is 24-36% faster on average for implicit linked list algorithms.