# 99 questions/Solutions/19

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< 99 questions | Solutions(Difference between revisions)

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rotate l n = rotate l (length l + n) | rotate l n = rotate l (length l + n) | ||

</haskell> | </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: | There are two separate cases: | ||

Line 18: | Line 20: | ||

rotate xs n = take len . drop (n `mod` len) . cycle $ xs | rotate xs n = take len . drop (n `mod` len) . cycle $ xs | ||

where len = length xs | where len = length xs | ||

+ | </haskell> | ||

+ | |||

+ | or without mod: | ||

+ | <haskell> | ||

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

</haskell> | </haskell> | ||

Line 30: | Line 37: | ||

or | 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> | <haskell> | ||

Line 35: | Line 48: | ||

where | where | ||

nn = n `mod` length xs | 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> | </haskell> |

## Revision as of 23:57, 19 November 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 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

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

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)

length

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)