Implement reverse in Haskell that runs in linear time

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I'm just learning Haskell, so sorry if my question is stupid. I'm reading learnyouahaskell.com and now I'm at chapter 5 "Recursion". There's an example of implementation of standard 'reverse' function:

reverse' :: [a] -> [a]  
reverse' [] = []  
reverse' (x:xs) = reverse' xs ++ [x]  

But it seems that it runs in O(N^2) time, while the standard reverse runs in O(N) (I hope so). The following code illustrates this:

sum (reverse [1,2..1000000]) -- runs pretty fast
sum (reverse' [1,2..1000000]) -- never finishes

So, I started thinking how to implement my own reverse faster. It's pretty easy to do in imperative languages. Maybe I need some more advanced material from subsequent chapters to do this? Any hints are welcomed.

4 Answers
foldl (\acc x -> x:acc) [] xs

This runs in O(n). The idea is pretty simple - you take an empty list(accumulator) and transfer elements to it from top to bottom.

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