Difference between revisions of "99 questions/Solutions/4"
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(Add description for each type of solution. Change formatting, ensure that all code is enclosed in "haskell" braces.) |
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(*) Find the number of elements of a list. |
(*) Find the number of elements of a list. |
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+ | The simple, recursive solution. |
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⚫ | |||
<haskell> |
<haskell> |
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myLength :: [a] -> Int |
myLength :: [a] -> Int |
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myLength [] = 0 |
myLength [] = 0 |
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myLength (_:xs) = 1 + myLength xs |
myLength (_:xs) = 1 + myLength xs |
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+ | </haskell> |
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⚫ | |||
+ | Same, but now we use an "accumulator" argument. |
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⚫ | |||
+ | <haskell> |
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⚫ | |||
⚫ | |||
⚫ | |||
where |
where |
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myLength_acc [] n = n |
myLength_acc [] n = n |
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</haskell> |
</haskell> |
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+ | Using foldl/foldr: |
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<haskell> |
<haskell> |
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− | myLength |
+ | myLength :: [a] -> Int |
− | + | myLength1 = foldl (\n _ -> n + 1) 0 |
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− | + | myLength2 = foldr (\_ n -> n + 1) 0 |
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− | + | myLength3 = foldr (\_ -> (+1)) 0 |
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− | + | myLength4 = foldr ((+) . (const 1)) 0 |
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− | + | myLength5 = foldr (const (+1)) 0 |
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+ | myLength6 = foldl (const . (+1)) 0 |
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</haskell> |
</haskell> |
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+ | We can also create an infinite list starting from 1. |
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+ | Then we "zip" the two lists together and take the last element (which is a pair) from the result: |
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<haskell> |
<haskell> |
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− | myLength |
+ | myLength :: [a] -> Int |
− | + | myLength1 xs = snd $ last $ zip xs [1..] -- Just for fun |
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− | + | myLength2 = snd . last . (flip zip [1..]) -- Because point-free is also fun |
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+ | myLength3 = fst . last . zip [1..] -- same, but easier |
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</haskell> |
</haskell> |
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+ | We can also change each element into our list into a '1' and then add them all together. |
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<haskell> |
<haskell> |
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⚫ | |||
myLength = sum . map (\_->1) |
myLength = sum . map (\_->1) |
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</haskell> |
</haskell> |
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− | |||
⚫ | |||
− | |||
− | -- length returns the length of a finite list as an Int. |
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− | length [] = 0 |
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− | length (_:l) = 1 + length l |
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− | |||
⚫ | |||
Revision as of 13:04, 15 May 2014
(*) Find the number of elements of a list.
The simple, recursive solution.
This is similar to the length
from Prelude
:
myLength :: [a] -> Int
myLength [] = 0
myLength (_:xs) = 1 + myLength xs
The prelude for haskell 2010 can be found here.
Same, but now we use an "accumulator" argument.
myLength :: [a] -> Int
myLength list = myLength_acc list 0
where
myLength_acc [] n = n
myLength_acc (_:xs) n = myLength_acc xs (n + 1)
Using foldl/foldr:
myLength :: [a] -> Int
myLength1 = foldl (\n _ -> n + 1) 0
myLength2 = foldr (\_ n -> n + 1) 0
myLength3 = foldr (\_ -> (+1)) 0
myLength4 = foldr ((+) . (const 1)) 0
myLength5 = foldr (const (+1)) 0
myLength6 = foldl (const . (+1)) 0
We can also create an infinite list starting from 1. Then we "zip" the two lists together and take the last element (which is a pair) from the result:
myLength :: [a] -> Int
myLength1 xs = snd $ last $ zip xs [1..] -- Just for fun
myLength2 = snd . last . (flip zip [1..]) -- Because point-free is also fun
myLength3 = fst . last . zip [1..] -- same, but easier
We can also change each element into our list into a '1' and then add them all together.
myLength :: [a] -> Int
myLength = sum . map (\_->1)