Difference between revisions of "99 questions/Solutions/17"
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(cleanup) |
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| Line 3: | Line 3: | ||
Do not use any predefined predicates. |
Do not use any predefined predicates. |
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| − | Solution using take and drop: |
+ | Solution using <hask>take</hask> and <hask>drop</hask>: |
| + | |||
<haskell> |
<haskell> |
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split xs n = (take n xs, drop n xs) |
split xs n = (take n xs, drop n xs) |
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</haskell> |
</haskell> |
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| + | Or even simpler using <hask>splitAt</hask>: |
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| ⚫ | |||
| + | |||
| + | <haskell> |
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| + | split = flip splitAt |
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| + | </haskell> |
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| + | |||
| ⚫ | |||
| + | |||
<haskell> |
<haskell> |
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split :: [a] -> Int -> ([a], [a]) |
split :: [a] -> Int -> ([a], [a]) |
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| Line 18: | Line 26: | ||
The same solution as above written more cleanly: |
The same solution as above written more cleanly: |
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| + | |||
<haskell> |
<haskell> |
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split :: [a] -> Int -> ([a], [a]) |
split :: [a] -> Int -> ([a], [a]) |
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| Line 23: | Line 32: | ||
split (x:xs) n = let (f,l) = split xs (n-1) in (x : f, l) |
split (x:xs) n = let (f,l) = split xs (n-1) in (x : f, l) |
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</haskell> |
</haskell> |
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| − | |||
| − | Note that this function, with the parameters in the other order, exists as <hask>splitAt</hask>. |
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Revision as of 20:13, 15 July 2010
(*) Split a list into two parts; the length of the first part is given.
Do not use any predefined predicates.
Solution using take and drop:
split xs n = (take n xs, drop n xs)
Or even simpler using splitAt:
split = flip splitAt
But these should clearly be considered "predefined predicates". Alternatively, we have the following recursive solution:
split :: [a] -> Int -> ([a], [a])
split [] _ = ([], [])
split l@(x : xs) n | n > 0 = (x : ys, zs)
| otherwise = ([], l)
where (ys,zs) = split xs (n - 1)
The same solution as above written more cleanly:
split :: [a] -> Int -> ([a], [a])
split xs 0 = ([], xs)
split (x:xs) n = let (f,l) = split xs (n-1) in (x : f, l)