# Difference between revisions of "99 questions/Solutions/15"

From HaskellWiki

< 99 questions | Solutions

(Added an alternative and more verbose solution) |
(discovered an interesting solution that was not on here) |
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repli' _ 0 = [] |
repli' _ 0 = [] |
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repli' x n = x : repli' x (n-1) |
repli' x n = x : repli' x (n-1) |
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+ | </haskell> |
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+ | |||

+ | or, a convoluted recursive solution that only uses cons: |
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+ | <haskell> |
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+ | repli :: [a] -> Int -> [a] |
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+ | repli [] _ = [] |
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+ | repli (x:xs) n = foldl (\f _ -> (x:) . f) (repli xs) [1..n] n |
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</haskell> |
</haskell> |

## Revision as of 09:42, 23 January 2012

(**) 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 convoluted recursive solution that only uses cons:

```
repli :: [a] -> Int -> [a]
repli [] _ = []
repli (x:xs) n = foldl (\f _ -> (x:) . f) (repli xs) [1..n] n
```