[Haskell-cafe] List of numbers to list of ranges

[Stephen Tetley]

I'd go with direct recursion for this one - the pattern of consumption
and production that generates the answer doesn't seem to neatly match
any of the standard recursion combinators (map, unfold, fold,
mapAccum, ...) nor exotic ones (skipping streams c.f. the Stream
fusion paper, apomorphisms, ...).

Direct recursion might be prosaic, but it is pleasantly obvious:

ranges :: (Num a, Eq a) => [a] -> [(a,a)]
ranges []     = []
ranges (a:as) = step (a,a) as
  where
    step (i,j) []                = [(i,j)]
    step (i,j) (x:xs) | j+1 == x = step (i,x) xs
    step (i,j) (x:xs)            = (i,j) : step (x,x) xs