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99 questions/Solutions/4

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(Added another two solutions)
Line 8: Line 8:
   
 
<haskell>
 
<haskell>
myLength' = foldl (\n _ -> n + 1) 0
+
myLength' = foldl (\n _ -> n + 1) 0
myLength'' = foldr (\_ n -> n + 1) 0
+
myLength'' = foldr (\_ n -> n + 1) 0
myLength''' = foldr (\_ -> (+1)) 0
+
myLength''' = foldr (\_ -> (+1)) 0
myLength'''' = foldr ((+) . (const 1)) 0
+
myLength'''' = foldr ((+) . (const 1)) 0
  +
myLength''''' = foldr (const (+1)) 0
 
</haskell>
 
</haskell>
   
 
<haskell>
 
<haskell>
 
myLength' xs = snd $ last $ zip xs [1..] -- Just for fun
 
myLength' xs = snd $ last $ zip xs [1..] -- Just for fun
  +
myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun
 
</haskell>
 
</haskell>
   

Revision as of 20:38, 21 November 2011

(*) Find the number of elements of a list.

myLength           :: [a] -> Int
myLength []        =  0
myLength (_:xs)    =  1 + myLength xs
myLength'     =  foldl (\n _ -> n + 1) 0
myLength''    =  foldr (\_ n -> n + 1) 0
myLength'''   =  foldr (\_ -> (+1)) 0
myLength''''  =  foldr ((+) . (const 1)) 0
myLength''''' =  foldr (const (+1)) 0
myLength' xs = snd $ last $ zip xs [1..] -- Just for fun
myLength''   = snd . last . (flip zip [1..]) -- Because point-free is also fun
myLength = sum . map (\x -> 1)
This is
length
in
Prelude
.