Good!

I came up with the same solution, thought I was doing some horrible hack xD

Perhaps, the best soultion to this problem.

I believe you'd do better in terms of time efficiency if you used a set to check which elements have already been added: checking membership of a list is a linear-time operation, but for a set it's pretty much constant (neglecting collisions).

But this is better in terms of memory efficiency, so I guess there are pros and cons.

Formatting the code would definitely make it more readable!