haskell - 返回一个列表，其中包含一对元素，但前提是各个元素的总和是奇数

Question

Implement the oddPairs :: [Int] -> [Int] -> [(Int, Int)] function that returns a list of pairs, but only if the parameters' lists' respective elements' sums are odd.

For example:

oddPairs [1,2,3] [2,2,2] == [(1,2),(3,2)]

oddPairs [1,3,5] [2,4,6] == zip [1,3,5] [2,4,6]

oddPairs [1,2,3] [1,2,3] == []

到目前为止，我已经尝试过

oddPairs (x:xs) (y:ys) | (x+y) `mod` 2 == 0 = []
                       | (x+y) `mod` 2 /= 0 = [(x, y)] ++ oddPairs (xs) (ys)

在第一个示例中，它仅返回 [(1,2)]，在第二个示例中，它返回正确的值，但具有 非穷举模式错误。

Answer 1

如果两个项目是偶数，你不应该只返回一个空列表，而是继续递归，直到至少有一个列表被用完，所以:

oddPairs :: Integral a => [a] -> [a] -> [(a, a)]
oddPairs [] _ = []
oddPairs _ [] = []
oddPairs (x:xs) (y:ys)
    -- keep searching for new items ↓
    | (x+y) `mod` 2 == 0 = <strong>oddPairs xs ys</strong>
    | otherwise = (x, y) : oddPairs xs ys