99 questions/Solutions/19
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m (new solution using replicate) |
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where | where | ||
nn = n `mod` length xs | nn = n `mod` length xs | ||
| + | </haskell> | ||
| + | |||
| + | or | ||
| + | |||
| + | <haskell> | ||
| + | rotate xs n = take len $ drop i $ concat $ replicate 2 xs | ||
| + | where len = length xs | ||
| + | i = n `mod` len | ||
</haskell> | </haskell> | ||
Revision as of 22:30, 15 August 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 = drop nn xs ++ take nn xs where nn = n `mod` length xs
or
rotate xs n = take len $ drop i $ concat $ replicate 2 xs where len = length xs i = n `mod` len
