给定一个二叉树,我试图找到元素出现次数最多的二叉树级别,例如Apple
。
以下是规格:
data Fruit = Peach | Apple
data BTree a = Empty | Node a (BTree a) (BTree a)
levelWithMaxApples :: BTree Fruit -> Int
我能够创建一个函数,给定一个二叉树,它将计算其中Apple
的出现次数。尽管如此,我还是无法弄清楚如何找到Apple
出现次数最多的水平,有什么提示吗?
以下是一些测试:
tree1 = Node Peach Empty Empty
tree2 = Node Peach
(Node Peach
(Node Apple
Empty
Empty)
(Node Peach
(Node Peach
Empty
Empty)
(Node Peach
Empty
Empty)))
(Node Apple
(Node Apple
Empty
(Node Peach
Empty
Empty))
Empty)
> levelWithMaxApples tree1
Nothing
> levelWithMaxApples tree2
2
这是我的尝试(在这种情况下,函数的名称应该countApples
:
levelWithMaxApples Empty = 0
levelWithMaxApples (Node Apple l r) = 1 + levelWithMaxApple l + levelWithMaxApple r
levelWithMaxApples (Node Peach l r) = levelWithMaxApple l + levelWithMaxApple r
感谢您的帮助!
这个答案是有文化的哈斯克尔。您可以使用.lhs
扩展名保存它并将其加载到 GHCi 中。
> import Data.Ord (comparing)
> import Data.List (maximumBy)
> data Fruit = Peach | Apple
> data BTree a = Empty | Node a (BTree a) (BTree a)
尝试将其分解成更小的部分。首先,编写一个函数,该函数将生成树中所有级别的列表:
> levels :: BTree a -> [[a]]
> levels Empty = []
> levels (Node x l r) = [x] : combine (levels l) (levels r)
> where combine [] ys = ys
> combine xs [] = xs
> combine (x:xs) (y:ys) = (x ++ y) : combine xs ys
(请注意,这里的combine
辅助函数就像zipWith (++)
一样,但在最短输入用尽后继续(。
一旦有了它,就很容易在列表中找到Apple
的实例(如果你在Fruit
定义中添加deriving Eq
,会更容易(:
> countEach :: (a -> Bool) -> [[a]] -> [Int]
> countEach pred = map (length . filter pred)
> countApples :: [[Fruit]] -> [Int]
> countApples = countEach isApple
> where isApple Apple = True
> isApple _ = False
接下来,您可以使用 zip
简单地使用索引号标记列表中的每个项目,然后使用 maximumBy
选择计数最大的项目:
> levelWithMaxApples :: BTree Fruit -> Int
> levelWithMaxApples t = let ls = levels t
> counts = countApples ls
> labeled = zip [0..] counts
> in fst . maximumBy (comparing snd) $ labeled
这是另一种方式。
-
从帮助程序函数开始。这将简单地构建该节点的原始列表和
Apple
计数。但是,这是为每个子树中的每个节点单独完成的,即列表将类似于[(根,1(,(L1,1(,(L2,0(,(R1,1(]import qualified Data.Map as M (fromListWith,toList) import qualified Data.List as L (sortBy) countApples' :: BTree Fruit -> Int -> [(Int,Int)] countApples' Empty _ = [] countApples' (Node Apple l r) n = (n,1) : (countApples' l (n+1)) ++ (countApples' r (n+1)) countApples' (Node Peach l r) n = (n,0) : (countApples' l (n+1)) ++ (countApples' r (n+1))
-
接下来,我们使用上述步骤中的内容创建一个地图。在此步骤中,我们将聚合树的每个级别的值。然后我们转换回列表,并按元组的值部分降序排序。然后,具有最大苹果实例的级别将是列表顶部元组中的第一个元素。
levelWithMaxApples :: BTree Fruit -> Int levelWithMaxApples Empty = error "Empty tree" levelWithMaxApples x = fst $ head $ L.sortBy ((k1,v1) (k2,v2) -> compare v2 v1) $ M.toList $ M.fromListWith (+) $ countApples' x 0
演示
但是,您应该注意,这可能不是一个非常有效的解决方案,因为除了排序之外,还要与 Map 进行转换。
注意:排序部分主要基于此答案。
更新:这是一种完全使用地图的方法。
import qualified Data.Map as M
countApples :: BTree Fruit -> Int -> M.Map Int Int
countApples Empty _ = M.empty
countApples (Node Apple l r) n = M.unionsWith (+) [(M.singleton n 1 ),(countApples l (n+1)),(countApples r (n+1))]
countApples (Node Peach l r) n = M.unionsWith (+) [(M.singleton n 0 ),(countApples l (n+1)),(countApples r (n+1))]
levelWithMaxApples :: BTree Fruit -> Int
levelWithMaxApples t = fst $ M.foldWithKey (k v acc@(k',v') -> if v >= v' then (k,v) else acc) (-1,-1) $ countApples t 0
演示