View source for 99 questions/Solutions/19

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

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

<haskell>
rotate [] _ = []
rotate xs 0 = xs
rotate (x:xs) (n+1) = rotate (xs ++ [x]) n
rotate xs n = rotate xs (length xs + n)
</haskell>

(Note that this solution uses [http://en.wikibooks.org/wiki/Haskell/Pattern_matching#n.2Bk_patterns n+k-patterns] which are [http://www.haskell.org/onlinereport/haskell2010/haskellli2.html#x3-5000 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:
<haskell>
rotate xs n = take len . drop (n `mod` len) . cycle $ xs
    where len = length xs
</haskell>

or using list comprehension:
<haskell>
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)
</haskell>

or without mod:
<haskell>
rotate xs n = take (length xs) $ drop (length xs + n) $ cycle xs
</haskell>

or

<haskell>
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
</haskell>

or

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

<haskell>
rotate xs n = drop nn xs ++ take nn xs
    where 
      nn = n `mod` length xs
</haskell>

Using a simple splitAt trick
<haskell>
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
</haskell>

Without using <hask>length</hask>:
<haskell>
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)
</haskell>

A much simpler solution without using <hask>length</hask> that is very similar to the first solution:
<haskell>
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)
</haskell>


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