我想从int列表[Int] -> [T]
生成所有可能的树,但我只生成一棵树。
1 1
2 2
3 5
4 14
5 42
像这些加泰罗尼亚数字一样。如果我的列表大小为3,我想生成5个可能的树,如果4到14个可能的树。
代码:
data T = N T T | L Int deriving (Show)
toT :: [Int] -> T
toT [] = L 0
toT [n] = L n
toT ns = T (toT (take mid ns)) (toT (drop (mid+1) ns))
where
mid = length ns div 2
例如:toT [1..3]
输出:N (L 1) (N (L 2) (L 3))
和N (N (L 1) (L 2)) (L 3)
。
现在ı确实喜欢这个
toTree [] = error "!!"
toTree [n] = Leaf n
toTree ns = Node leftTree rightTree
where
leftTree = toTree $ take (length(ns)-1) ns
rightTree = toTree $ drop (length(ns)-1) ns` ı want ns length contiue descend one point recursive but ı didnt
怎么能这样做?在recursiveı将发送相同的列表,但长度将下降ı发送[1,2,3]大小3再次发送[1,2,3]长度2
答案 0 :(得分:1)
需要发生的是,列表需要在每个可能的位置进行拆分,这样做会使列表变小。需要累积子列表中的树,然后将所有内容组合在一起。
data Tree
= L {-# UNPACK #-} !Int
| N !Tree !Tree
deriving (Eq, Ord, Read, Show)
-- Convert a list of Ints into a list of Trees containing the given list.
toTrees :: [Int] -> [Tree]
-- We start with the base cases: 0 and 1 elements. Because there are no
-- trees of 0 length, it returns the empty list in that case.
toTrees [] = []
toTrees [x] = [L x]
-- There is at least two elements in this list, so the split into nonempty
-- lists contains at least one element.
toTrees (x:xs@(y:ys)) = let
-- splitWith uses a difference list to accumulate the left end of the
-- split list.
splitWith :: ([a] -> [a]) -> [a] -> [([a], [a])]
splitWith fn [] = []
splitWith fn as@(a:as') = (fn [], as):splitWith (fn . (:) a) as'
-- Now we use a list comprehension to take the list of trees from each
-- split sublist.
in [
N tl tr |
(ll, lr) <- ([x], xs):splitWith ((:) x . (:) y) ys,
tl <- toTrees ll,
tr <- toTrees lr
]
这给出了期望的结果:
GHCi> toTrees [1, 2, 3, 4, 5]
[N (L 1) (N (L 2) (N (L 3) (N (L 4) (L 5)))),N (L 1) (N (L 2) (N (N (L 3) (L 4)) (L 5))),
N (L 1) (N (N (L 2) (L 3)) (N (L 4) (L 5))),N (L 1) (N (N (L 2) (N (L 3) (L 4))) (L 5)),
N (L 1) (N (N (N (L 2) (L 3)) (L 4)) (L 5)),N (N (L 1) (L 2)) (N (L 3) (N (L 4) (L 5))),
N (N (L 1) (L 2)) (N (N (L 3) (L 4)) (L 5)),N (N (L 1) (N (L 2) (L 3))) (N (L 4) (L 5)),
N (N (N (L 1) (L 2)) (L 3)) (N (L 4) (L 5)),N (N (L 1) (N (L 2) (N (L 3) (L 4)))) (L 5),
N (N (L 1) (N (N (L 2) (L 3)) (L 4))) (L 5),N (N (N (L 1) (L 2)) (N (L 3) (L 4))) (L 5),
N (N (N (L 1) (N (L 2) (L 3))) (L 4)) (L 5),N (N (N (N (L 1) (L 2)) (L 3)) (L 4)) (L 5)]
GHCi> length it
14
答案 1 :(得分:-2)
我更新了你的代码,它似乎对我有用。你能检查一下这是否符合你的期望吗?
import Data.List
data Tree = Node Tree Tree | Leaf Int deriving (Show)
toTree :: [Int] -> Tree
toTree [] = Leaf 0
toTree [n] = Leaf n
toTree ns = Node leftTree rightTree
where midIndex = (length ns) `div` 2
leftTree = toTree $ take midIndex ns
rightTree = toTree $ drop (midIndex+1) ns
allTrees :: [Int] -> [Tree]
allTrees xs = map toTree $ permutations xs
main = print $ allTrees [1..4]