在haskell中。如何将拆分列表解析为元组

Question

splitList n xs：基于提供的索引n将列表xs拆分为两个列表的元组。

例子：

splitList 3 [1..5] ⇒ ([1,2,3], [4,5]) 
splitList 3 [1..] ⇒ ([1,2,3], [4..]) 
splitList 9 [] ⇒ ([], []) 

如何解决这个问题呢？

Answer 1

这是一个可能的解决方案：

splitList :: Int -> [a] -> ([a], [a]) 
splitList _ [] = ([], []) 
splitList n xs = (take n xs, drop n xs) 

Answer 2

splitAt :: Int -> [a] -> ([a], [a]) 

λ> splitAt 3 [1..5] 
([1,2,3],[4,5])