99 questions/Solutions/19

(**) Rotate a list N places to the left.

Hint: Use the predefined functions length and (++).

rotate [] _ = []
rotate xs 0 = xs
rotate (x:xs) (n+1) = rotate (xs ++ [x]) n
rotate xs n = rotate xs (length xs + n)

(Note that this solution uses n+k-patterns which are removed from Haskell 2010.)

There are two separate cases:

• If n > 0, move the first element to the end of the list n times.
• If n < 0, convert the problem to the equivalent problem for n > 0 by adding the list's length to n.

or using cycle:

rotate xs n = take len . drop (n `mod` len) . cycle \$ xs
where len = length xs

or using list comprehension (only works for sequential increasing elements):

rotate :: (Enum a) => [a] -> Int -> [a]
rotate xs n = [(f n) .. last xs] ++ [head xs .. (f (n-1))]
where f k = xs !! (k `mod` length xs)

or without mod:

rotate xs n = take (length xs) \$ drop (length xs + n) \$ cycle xs

or

rotate xs n = if n >= 0 then
drop n xs ++ take n xs
else let l = ((length xs) + n) in
drop l xs ++ take l xs

or

rotate xs n | n >= 0 = drop n xs ++ take n xs
| n < 0 = drop len xs ++ take len xs
where len = n+length xs

or calculate the position(index) at first:

rotate xs n = let i = if n < 0 then length xs + n else n
in drop i xs ++ take i xs

or

rotate xs n = drop nn xs ++ take nn xs
where
nn = n `mod` length xs

Using a simple splitAt trick

rotate xs n
| n < 0 = rotate xs (n+len)
| n > len = rotate xs (n-len)
| otherwise = let (f,s) = splitAt n xs in s ++ f
where len = length xs

Without using length:

rotate xs n
| n > 0 = (reverse . take n . reverse \$ xs) ++ (reverse . drop n . reverse \$ xs)
| n <= 0 = (drop (negate n) xs) ++ (take (negate n) xs)

A much simpler solution without using length that is very similar to the first solution:

rotate :: [a] -> Int -> [a]
rotate [] _ = []
rotate x 0 = x
rotate x y
| y > 0 = rotate (tail x ++ [head x]) (y-1)
| otherwise = rotate (last x : init x) (y+1)