The "alternative solution" posted there doesn't suffer from the monomorphism restriction*
, and thus does less to propagate the myth that one has to be a genius to write code in Haskell. It also works correctly with empty input :-)
*Something that, for the casual programmer, is quite technical but uninteresting, has a scary name, and a not-that-friendly error message.
compress  = 
compress [a] = [a] -- singleton list
compress (x:y:xs) = (if x == y then  else [x]) ++ compress (y:xs)
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