Difference between revisions of "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) |
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</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.) |
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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)