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99 questions/Solutions/19

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(Add note to first solution that n+k patterns are discouraged)
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rotate l n = rotate l (length l + n)
 
rotate l n = rotate l (length l + n)
 
</haskell>
 
</haskell>
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(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:
 
There are two separate cases:

Revision as of 14:00, 6 February 2011

(**) 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)

(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

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
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)