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

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< 99 questions | Solutions

(discovered an interesting solution that was not on here) |
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− | or, a convoluted recursive solution that only uses cons: |
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+ | or, a version that does not use list concatenation: |
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<haskell> |
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repli :: [a] -> Int -> [a] |
repli :: [a] -> Int -> [a] |
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repli [] _ = [] |
repli [] _ = [] |
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− | repli (x:xs) n = |
+ | repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n] |

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+ | [[Category:Programming exercise spoilers]] |

## Latest revision as of 19:33, 18 January 2014

(**) 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]
```