Difference between revisions of "99 questions/Solutions/2"
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(Add foldl based solution.) |
m (Add description to foldl solution.) |
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myButLast'''' = head . tail . reverse |
myButLast'''' = head . tail . reverse |
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| + | </haskell> |
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| + | We can also use a pair to keep the last two elements processed from the list. We take advantage of lazy evaluation, such that if the list is too small, the initial elements in the pair get evaluated, resulting in an error: |
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| + | <haskell> |
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myButLast''''' = snd.(foldl (\(a,b) c -> (c,a)) (e1, e2)) |
myButLast''''' = snd.(foldl (\(a,b) c -> (c,a)) (e1, e2)) |
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where e1 = error "List too small!" |
where e1 = error "List too small!" |
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Revision as of 15:27, 15 May 2014
(*) Find the last but one element of a list.
myButLast :: [a] -> a
myButLast = last . init
myButLast' x = reverse x !! 1
myButLast'' [x,_] = x
myButLast'' (_:xs) = myButLast'' xs
myButLast''' (x:(_:[])) = x
myButLast''' (_:xs) = myButLast''' xs
myButLast'''' = head . tail . reverse
We can also use a pair to keep the last two elements processed from the list. We take advantage of lazy evaluation, such that if the list is too small, the initial elements in the pair get evaluated, resulting in an error:
myButLast''''' = snd.(foldl (\(a,b) c -> (c,a)) (e1, e2))
where e1 = error "List too small!"
e2 = error "List is null!"