Haskell:将偶数和奇数元素拆分为元组 [英] Haskell: split even and odd elements into tuple

问题描述

我不能使用高阶函数.我只是看不出如何做到这一点.我对Haskell非常陌生.它也必须是递归的.

I can't use high order functions. I just can't see to figure out how to do this. I am very new to haskell. It also has to be recursive.

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

我首先得到这个.老实说，我什至不知道从哪里开始这个问题.

I am given this to start with. I honestly don't even know where to start with this problem.

示例:

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])

任何帮助将不胜感激.

其偶数和奇数位置.

所以

分割[3,6,8,9,10]将是 ([3,8,10]，[6,9])

split [3,6,8,9,10] would be ([3,8,10],[6,9])

好的，所以我想到了这个.它不漂亮，但是看起来还可以.

ok so i came up with this. Its not pretty, but it seems to work ok.

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)))

推荐答案

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

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