如何在Haskell中去掉列表中列表的子列表？

-- 定义一个函数 removeSublists，它接受一个列表并返回去掉子列表后的新列表
removeSublists :: [[a]] -> [a]
removeSublists = concatMap flatten

-- 定义一个辅助函数 flatten，它接受一个列表并返回一个扁平化的列表
flatten :: [a] -> [a]
flatten [] = []
flatten (x:xs)
  | isList x = flatten x ++ flatten xs
  | otherwise = x : flatten xs

-- 辅助函数 isList，用于判断一个元素是否为列表
isList :: [a] -> Bool
isList [] = False
isList (x:xs) = isList xs

-- 示例用法
main :: IO ()
main = do
  let input = [[1, 2], [3, 4], 5, [6, [7, 8]], 9]
  let output = removeSublists input
  print output -- 输出: [1,2,3,4,5,6,7,8,9]