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haskell:将偶数和奇数元素拆分为元组

haskell

3

Matt · 技术社区 · 16 年前

我不能用高阶函数。我就是不知道该怎么做。我对哈斯克尔很陌生。它还必须是递归的。

split :: [Int] -> ([Int],[Int])
split xs =

我要从这个开始。老实说,我甚至不知道从哪里开始这个问题。

实例:

split []
([],[])

split [1]
([1],[])

split [1,2,3,4,5,6,7,8,9,10]
([1,3,5,7,9],[2,4,6,8,10])

任何帮助都将不胜感激。

编辑:它的偶数和奇数位置。

所以

拆分[3,6,8,9,10]将是 ([3,8,10],[6,9])

好吧,所以我想到了这个。它不漂亮,但似乎工作正常。

split :: [Int] -> ([Int],[Int])
split [] = ([],[])
split [xs] = ([xs],[])
split xs = (oddlist xs, evenlist xs)

oddlist :: [Int] -> ([Int])
oddlist xs | length xs <= 2 = [head(xs)]
           | otherwise = [head(xs)] ++ oddlist(tail(tail(xs)))

evenlist :: [Int] -> ([Int])
evenlist xs | length xs <= 3 = [head(tail(xs))]
            | otherwise = [head(tail(xs))] ++ evenlist(tail(tail(xs)))

5 回复 | 直到 10 年前

1

17

kennytm 16 年前

split [] = ([], [])
split [x] = ([x], [])
split (x:y:xs) = (x:xp, y:yp) where (xp, yp) = split xs

2

5

dfeuer 10 年前

split :: [a] -> ([a],[a])
split = foldr (\x ~(y2,y1) -> (x:y1, y2)) ([],[])

~ split

foldr

split :: [a] -> ([a],[a])
split [] = ([],[])
split (x : xs) = (x : y1, y2)
  where
    (y2, y1) = split xs

3

Tom Lokhorst 16 年前

-- Base case:
split [] = â¦

-- Recurrence:
split (x : xs) = (do something with x) â¦Â (split xs) â¦

4

2

Matt Ellen Bipin Vayalu 16 年前

split xs = (everyother 0 xs, everyother 1 xs)
      where everyother _ []     = []
            everyother 1 (x:xs) = everyother 0 xs
            everyother 0 (x:xs) = x : (everyother 1 xs)

5

1

Community Mohan Dere 9 年前

Get every Nth element

ghci> let split ys = let skip xs = case xs of { [] -> [] ; [x] -> [x] ; (x:_:xs') -> x : skip xs' } in (skip ys, skip . drop 1 $ ys)
ghci> split [1..10]
([1,3,5,7,9],[2,4,6,8,10])

split xs = (skip xs, skip . drop 1 $ xs)
  where 
  skip [] = []
  skip [x] = [x]
  skip (x:_:xs') = x : skip xs'