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99 questions/Solutions/19

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

Revision as of 19:30, 6 December 2010

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

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

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

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 

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
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
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