如何在列表列表上获得递归功能的输出，如何获得多个列表？

Question

我正在尝试使用一个递归功能，该功能打印所有具有最大长度列表中最大长度的列表（如果它们的长度相同，则可能是一个或多个）

给定输入，例如：

[[],[3],[2],[6],[3,6],[2,6],[4],[3,4],[2,4],[5],[3,5],[2,5],[4,5],[3,4,5],[2,4,5],[1]]

输出应包含两个最长列表

[3,4,5],[2,4,5]

：以下功能仅打印第一个列表：[3,4,5]

  longest :: Ord a => [[a]] -> [a]
  longest [y] = y    --base case: if there's only one element left, return it.
  longest (x:y:lst) --extract the first two elements x, y from the list.  
     | length x < length y = longest (y:lst)
     | otherwise  = longest (x:lst)

注意：我必须使用恢复

Answer 1

您可以使用累加器跟踪迄今为止获得的最长项目以及此列表的列表，因此：

longest :: [[a]] -> [[a]]
longest (x:xs) = go xs (length x) [x]
    go [] _ ys = reverse ys
    go (x:xs) n ys
        | n' > n = go xs n' [x]
        | n' == n = go xs n (x:ys)
        | otherwise = go xs n ys
        where n' = length x

Answer 2

我发现您的方法很复杂，因为您将在参数开始时积累结果，并且有必要进一步使用。
考虑此解决方案，该解决方案将未来结果累积到第二个辅助参数中。

longest :: [[a]] -> [[a]]
longest lst = longest' lst [[]]

longest' :: [[a]]->[[a]]->[[a]] -- input, working state, output
longest' [] x = x --base case: if there's empty input return working state.
longest' (y:lst) x
 | length (head x) < length y = longest' lst [y]
 | length (head x) == length y = longest' lst (x++[y])
 | otherwise = longest' lst x

inp = [[],[3],[2],[6],[3,6],[2,6],[4],[3,4],[2,4],[5],[3,5],[2,5],[4,5],[3,4,5],[2,4,5],[1]]

main = putStrLn $ show $ longest inp

输出：

[[3,4,5],[2,4,5]]

您可以在此设计中的SO或标准库中在Haskell中看到这种方法：

longest lst = longest' lst [[]]
  where 
    longest' [] x = x --base case: if there's empty input return helper.
    longest' (y:lst) x
     | length (head x) < length y = longest' lst [y]
     | length (head x) == length y = longest' lst (x++[y])
     | otherwise = longest' lst x