There are several ways to solve this problem in Haskell. The naive approach would be to use the concatenate function ++
:
reverseList [] = []
reverseList (x:xs) = reverseList xs ++ [x]
However, this will be really slow for large lists since Haskell lists are really singly linked lists, so in order to append an element you have to traverse the entire list. An alternative would be to keep up with the list you're building in a helper function:
reverseList = go []
where
go acc [] = acc
go acc (x:xs) = go (x:acc) xs
However, this is really just the fold
pattern:
reverseList = foldl (acc x -> x : acc) []
But acc x -> x : acc
is just flip (:)
, so this can be written as
reverseList = foldl (flip (:)) []
However, the easiest way would probably be to just use the reverse
function in Prelude.
I would like to point out that your type of reverseList :: [Int] -> [Int]
could be generalized to :: [a] -> [a]
, you don't do anything special with the elements of the list, you're just building a new list with them.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…