Need to partition a list into lists based on breaks in ascending order of elements (Haskell)

Question

Say I have any list like this:

[4,5,6,7,1,2,3,4,5,6,1,2]

I need a Haskell function that will transform this list into a list of lists which

Answer 1

You can use a right fold to break up the list at down-steps:

foldr foo [] xs
  where
    foo x yss = (x:zs) : ws
      where
        (zs, ws) = case yss of
                     (ys@(y:_)) : rest
                            | x < y     -> (ys,rest)
                            | otherwise -> ([],yss)
                     _ -> ([],[])

(It's a bit complicated in order to have the combining function lazy in the second argument, so that it works well for infinite lists too.)