Difference between revisions of "99 questions/Solutions/19"
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Blazedaces (talk | contribs) (Added another solution) |
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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: |
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rotate xs n = take len . drop (n `mod` len) . cycle $ xs |
rotate xs n = take len . drop (n `mod` len) . cycle $ xs |
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where len = length xs |
where len = length xs |
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+ | </haskell> |
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+ | |||
+ | or without mod: |
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+ | <haskell> |
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⚫ | |||
</haskell> |
</haskell> |
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or |
or |
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+ | |||
+ | <haskell> |
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+ | rotate xs n | n >= 0 = drop n xs ++ take n xs |
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⚫ | |||
⚫ | |||
+ | </haskell> |
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<haskell> |
<haskell> |
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</haskell> |
</haskell> |
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+ | Using a simple splitAt trick |
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− | or |
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+ | <haskell> |
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+ | rotate xs n |
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+ | | n < 0 = rotate xs (n+len) |
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+ | | n > len = rotate xs (n-len) |
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+ | | otherwise = let (f,s) = splitAt n xs in s ++ f |
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+ | where len = length xs |
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+ | </haskell> |
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+ | |||
+ | Without using <hask>length</hask>: |
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+ | <haskell> |
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+ | rotate xs n |
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+ | | n > 0 = (reverse . take n . reverse $ xs) ++ (reverse . drop n . reverse $ xs) |
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+ | | n <= 0 = (drop (negate n) xs) ++ (take (negate n) xs) |
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+ | </haskell> |
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+ | A much simpler solution without using <hask>length</hask> that is very similar to the first solution: |
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<haskell> |
<haskell> |
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+ | rotate :: [a] -> Int -> [a] |
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⚫ | |||
+ | rotate [] _ = [] |
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⚫ | |||
+ | rotate x 0 = x |
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⚫ | |||
+ | rotate x y |
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+ | | y > 0 = rotate (tail x ++ [head x]) (y-1) |
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+ | | otherwise = rotate (last x : init x) (y+1) |
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</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
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)
A much simpler solution without using length
that is very similar to the first solution:
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)