haskell的一半功能

Question

使用庫函數，定義一個函數halve :: [a]→（[a]，[a]） 將一個偶數長度的列表分成兩半。例如：

> halve [1, 2, 3, 4, 5, 6] 
([1, 2, 3], [4, 5, 6]) 

到目前爲止我有什麼是

halve :: [a] -> ([a],[a]) 
halve = (\xs -> case xs of 
     [] -> ([],[]) 
     xs -> take ((length xs) `div` 2) xs) 

，它是錯誤的，因爲XS - >取（（長x）div 2）僅XS顯示列表的前半部分..請幫助我繼續下去，以便它顯示列表的後半部分。

Answer 1

感謝評論一些解決方案。我解決它......這裏是

first_halve :: [a] -> [a] 
first_halve = (\xs -> case xs of 
      [] -> [] 
      xs -> take ((length xs) `div` 2) xs) 

second_halve :: [a] -> [a] 
second_halve = (\xs -> case xs of 
      [] -> [] 
      xs -> drop ((length xs) `div` 2) xs) 

halve :: [a] -> ([a],[a]) 
halve = (\xs -> case xs of 
      [] -> ([],[]) 
      xs -> (first_halve xs, second_halve xs))