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(Added if-based solution)
 
Line 29: Line 29:
 
if length xs == 1 then x
 
if length xs == 1 then x
 
else myButLast''''' xs
 
else myButLast''''' xs
 +
 +
myButLast'''''' = head . reverse . init
 
</haskell>
 
</haskell>
  
  
 
[[Category:Programming exercise spoilers]]
 
[[Category:Programming exercise spoilers]]

Latest revision as of 14:30, 25 October 2017

(*) 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
 
lastbut1 :: Foldable f => f a -> a
lastbut1 = fst . foldl (\(a,b) x -> (b,x)) (err1,err2)
  where
    err1 = error "lastbut1: Empty list"
    err2 = error "lastbut1: Singleton"
 
lastbut1safe :: Foldable f => f a -> Maybe a
lastbut1safe = fst . foldl (\(a,b) x -> (b,Just x)) (Nothing,Nothing)
 
myButLast''''' [] = error "Empty list"
myButLast''''' [x] = error "Too few elements"
myButLast''''' (x:xs) = 
		if length xs == 1 then x
		else myButLast''''' xs
 
myButLast'''''' = head . reverse . init