# Difference between revisions of "99 questions/Solutions/19"

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< 99 questions | Solutions

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where |
where |
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nn = n `mod` length xs |
nn = n `mod` length xs |
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+ | </haskell> |
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+ | |||

+ | Using a simple splitAt trick |
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+ | <haskell> |
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+ | rotate xs n |
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+ | | n < 0 = rotate xs (n+len) |
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+ | | n > len = rotate xs (n-len) |
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+ | | otherwise = let (f,s) = splitAt n xs in s ++ f |
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+ | where len = length xs |
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</haskell> |
</haskell> |

## Revision as of 04:54, 20 November 2010

(**) Rotate a list N places to the left.

Hint: Use the predefined functions length and (++).

```
rotate [] _ = []
rotate l 0 = l
rotate (x:xs) (n+1) = rotate (xs ++ [x]) n
rotate l n = rotate l (length l + n)
```

There are two separate cases:

- If n > 0, move the first element to the end of the list n times.
- If n < 0, convert the problem to the equivalent problem for n > 0 by adding the list's length to n.

or using cycle:

```
rotate xs n = take len . drop (n `mod` len) . cycle $ xs
where len = length xs
```

or

```
rotate xs n = if n >= 0 then
drop n xs ++ take n xs
else let l = ((length xs) + n) in
drop l xs ++ take l xs
```

or

```
rotate xs n = drop nn xs ++ take nn xs
where
nn = n `mod` length xs
```

Using a simple splitAt trick

```
rotate xs n
| n < 0 = rotate xs (n+len)
| n > len = rotate xs (n-len)
| otherwise = let (f,s) = splitAt n xs in s ++ f
where len = length xs
```