99 questions/Solutions/15

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(**) Replicate the elements of a list a given number of times.

repli :: [a] -> Int -> [a]
repli xs n = concatMap (replicate n) xs

or, in Pointfree style:

repli = flip $ concatMap . replicate

alternatively, without using the

replicate

function:

repli :: [a] -> Int -> [a]
repli xs n = concatMap (take n . repeat) xs

or, using the list monad:

repli :: [a] -> Int -> [a]
repli xs n = xs >>= replicate n

or, a more verbose solution without the use of

replicate

repli :: [a] -> Int -> [a]
repli xs n = foldl (\acc e -> acc ++ repli' e n) [] xs
    where
      repli' _ 0 = []
      repli' x n = x : repli' x (n-1)

or, a version that does not use list concatenation:

repli :: [a] -> Int -> [a]
repli [] _ = []
repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n]

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99 questions/Solutions/15

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@@ Line 6: / Line 6: @@
 </haskell>
-or, in Pointfree style:
+or, in [[Pointfree]] style:
 <haskell>
 repli = flip $ concatMap . replicate
+</haskell>
+alternatively, without using the <hask>replicate</hask> function:
+<haskell>
+repli :: [a] -> Int -> [a]
+repli xs n = concatMap (take n . repeat) xs
+</haskell>
+or, using the list monad:
+<haskell>
+repli :: [a] -> Int -> [a]
+repli xs n = xs >>= replicate n
+</haskell>
+or, a more verbose solution without the use of <hask>replicate</hask>:
+<haskell>
+repli :: [a] -> Int -> [a]
+repli xs n = foldl (\acc e -> acc ++ repli' e n) [] xs
+    where
+      repli' _ 0 = []
+      repli' x n = x : repli' x (n-1)
+</haskell>
+or, a version that does not use list concatenation:
+<haskell>
+repli :: [a] -> Int -> [a]
+repli [] _ = []
+repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n]
 </haskell>