As part of my interview, I was asked to reverse a list. Fortunately or unfortunately, I had just read the JoelOnSoftware blog not to long ago, and he discussed asking candidates to reverse a list. So, I already had an idea in my mind of how to do it. It took me a while to get it, and my implementation was very awkward, but at least I finished. (In fact, I was told by the interviewer that he had never seen it implemented that way before.)
My implementation was recursive. That gets you the stack implicitly. I thought the point of the exercise was to see if you would recognize that you need a stack. But I later found out you can actually do it without one. This is the most efficient way to do it, but it's not the easiest to understand. The easiest to understand method is by using an explicit stack on the data. Of course, it's also the most inefficient.
Once I cleaned up my solution, the code was actually very small and
somewhat elegant. But the interface was awkward. You have to run the
following to complete the reversal.
The second method is similar, but the code is longer, and the interface is nicer. The third method doesn't need a stack. It's not pretty, but it works. The fourth method uses an explicit stack to reverse all the pointers. The fifth method does the same but reverses the data instead of the pointers. Of course, you could also implement an O(n^2) algorithm by traversing the list and reversing only the last non-reversed pointer each time. That would be about the worst way you could do it.
I also timed the reversals. The first reason I did this is because you really can't know performance without testing. The second reason is because my first interviewer wanted me to implement a tree traversal iteratively. This would be a lot easier for me to do recursively, but he cited stack overflow as being a problem. Again, I thought the point was to see if you could recognize that you need a stack.
Unfortunately, using an explicit stack that allocates memory from the heap is much slower than using the stack frame. It requires several extra function invocations (push, pop, malloc, free). And memory allocation and deallocation can be very slow. When you use the stack frame, there is hardly any work to "allocate" memory. And if you have a smart compiler and you get into a tail recursion situation, there's virtually no work to be done at all. Perhaps we should all just disabuse ourselves of this mutation fixation, and drop the heap altogether.
In any case, the results of the timing were exactly as expected.
I ran this serveral times. Reverse1 varied from 11 to 13 seconds. Reverse2 once came in at 16 seconds. The fastest method was stackless. The second fastest method used the stack frame. The slowest used a stack data structure.
I want to cover the stackless version of reverse (reverse3) since it's a bit tricky. You have to keep three temporary variables around so you don't lose your next pointers. I called them next, nextnext, and tmp, which is really nextnextnext. I trace through the iteration one time including the initial steps.
//setup head -> item1 -> item2 -> item3 -> NULL next^ nextnext^ //head->next = NULL; head -> item1 item2 -> item3 -> NULL NULL <--| next^ nextnext^ //tmp = nextnext->next; head -> item1 item2 -> item3 -> NULL NULL <--| next^ nextnext^ tmp^ //next->next = head; head -> item1 <- item2 item3 -> NULL NULL <--| next^ nextnext^ tmp^ //head = next; NULL <- item1 <- item2 item3 -> NULL head,next^ nextnext^ tmp^ //next = nextnext; NULL <- item1 <- item2 item3 -> NULL head^ next,nextnext^ tmp^ //nextnext = tmp; NULL <- item1 <- item2 item3 -> NULL head^ next^ nextnext,tmp^
In this example, the loop would only exectue one time. So the next step would point item3 back to item2 and return a pointer to item3.
And here it is in elisp.
(defun reverse (l) (if (eq l nil) () (append (reverse (cdr l)) (list (car l))))) (reverse '(1 2 3 4 5 6 7 8))
And in Python. This is why I love Python! I don't recommend you try this during an interview. Note that this is in place.
l = [1,2,3,4,5,6,7,8] l.reverse()
If you're going to be doing any technical interviews, you may want to study reversing a list since it seems to be a popular question. You will need to be very comfortable with pointers to understand my code. Also notice my list structure is very general. I initially had it store an integer. But reverse4 needed to be able to push pointers onto the stack; so, I modified the list to point to generic data.
Finally, I just want to mention one thing about data structures. I used the list as both a list and a stack. In reality they're the same thing. I like to think of the data representation as separate from the abstract data type. The interface to the data representation is the abstract data type.
For example, you have a list with insert. I could implement push and pop of the stack as insert(list,0) and remove(list,0) respectively. If you add an end pointer to the list, you can also implement a queue. Enqueue would be append(list) (which is really just insert(list,END)), and dequeue would be remove(list,0). You see the data representation underneath is the same. It's just that the interface is a little different in each case. But it's easier to think about push and pop than insert if what you wanted was a stack.
You can also change the underlying data representation while keeping the interface the same. You see, the two are independent. Lists, stacks, and queues can all be implemented using an array, but there are some consequences to doing so.
Also, notice how I passed the stack to the functions that operate on it. You should get into the habit of coding in this object-oriented manner. If I had used global variables instead, I couldn't easily extend this code to work with threads. And that's something that's becoming very important as we move into the multicore world.