(Add a solution with foldr along with some explanation.)
Revision as of 21:28, 8 May 2014
Insert an element at a given position into a list.
insertAt :: a -> [a] -> Int -> [a] insertAt x xs (n+1) = let (ys,zs) = split xs n in ys++x:zs
insertAt :: a -> [a] -> Int -> [a] insertAt x ys 1 = x:ys insertAt x (y:ys) n = y:insertAt x ys (n-1)
As a note to the above solution - this presumes that the inserted argument will be a singleton type a inserted into a list [a]. The lisp example does not infer this intent. As a result, presuming the data to be inserted is likewise of type [a] (which we are tacitly inferring here to be String into String insertion), a solution is:
insertAt x xs n = take (n-1) xs ++ [x] ++ drop (n-1) xs
This solution, like many others in this quiz presumes counting element positions starts at 1, perhaps causing needless confusion.
A solution using foldl and a closure, also assumes lists are 1 indexed:
insertAt :: a -> [a] -> Int -> [a] insertAt el lst n = fst $ foldl helper (,1) lst where helper (acc,i) x = if i == n then (acc++[el,x],i+1) else (acc++[x],i+1)
The use of foldl impose the use of concatenation. With a foldr we can use (:) instead, which is faster (O(n) vs. O(n²) I guess ?). The use of zip [1..] doesn't seem to had overhead compared to the same solution with the index stored in the accumulator.
insertAt :: a -> [a] -> Int -> [a] insertAt elt lst pos = foldr concat'  $ zip [1..] lst where concat' (i, x) xs | i == pos = elt:x:xs | otherwise = x:xs
Compared to the simple recursive definition, the fold version visit every elements of the list, whereas we could just stop after insertion of the element.