Haskell 中的发牌

Question

我需要定义一个函数

deal :: [String] -> [String] -> [(String,String)]

它需要像这样工作：

deal ["Hercule","Ariadne"] ["Ace","Joker","Heart"] -- Outputs [("Ace","Hercule"),("Joker","Ariadne"),("Heart","Hercule")]
take 4 (deal ["a","b","c"] (map show [0..]))       -- Outputs [("0","a"),("1","b"),("2","c"),("3","a")]
deal ("you":(repeat "me")) ["1","2","3","4"]       -- Outputs [("1","you"),("2","me"),("3","me"),("4","me")]

而且，老实说，这是一个练习；建议我使用

zip

和

cycle

。

我已经尝试解决这个问题三个半小时，但未能找到解决方案。

问：您能给我一个如何解决这个问题的提示吗？

我最好的尝试

deal :: [String] -> [String] -> [(String, String)]
deal ps cs = dealHelper ps cs ps cs []
  where dealHelper [] [] _ _ result = result
        dealHelper ps [] _ _ result = dealHelper ps (take (length ps) cs) result
        dealHelper [] cs allps allcs result = dealHelper (take (length cs))

（这甚至无法编译。）

Answer 1

解决方案是：

deal :: [String] -> [String] -> [(String,String)]
deal ps cs = zip cs (cycle ps)