Question

基本上，我有一个函数可以通过几个条件检查[1..9]的每个排列。另外，我有一个函数，它检查列表的总和是否等于数字和切片列表的函数。问题是它返回错误Couldn't match type 'Int' with 'Char'。它指向列表，如果所有条件都为真，则必须返回哪个函数。这是代码本身：

intSumList :: Int -> [Int] -> Bool
intSumList _ [] = True
intSumList x (y:ys)
                    | x == sum (y:ys) = True
                    | otherwise =  False
slice :: Int -> Int -> [Int] -> [Int]
slice from to s = take (to - from + 1)(drop from s)

find :: (Int,Int,Int,Int) -> [Int]
find (x,z,y,s) = checkPerm (x,z,y,s) l
                 where
                   l = L.permutations [1..9]
                   checkPerm (_,_,_,_ ) [] = ""
                   checkPerm (x,z,y,s) (p:pz)
                      | intSumList x (slice 1 5 p) && intSumList z (slice 2 6 p) && 
                        intSumList y (slice 4 8 p) && intSumList s (slice 5 9 p) = p
                      | otherwise = permutations (x,z,y,s) pz

Answer 1

当前问题是checkPerm返回一个类型为[Char]的空列表，而不是[Int]，当它获取一个空列表作为其第二个参数时。

但是，checkPerm基本上是重新实现Data.List.find。

import qualified Data.List as DL
import Data.Maybe
import Control.Monad

find :: (Int,Int,Int,Int) -> [Int]
find (x,z,y,s) = fromMaybe [] (DL.find p l)
                 where makePred x lo hi p = intSumList x (slice lo hi p)
                       p1 = makePred x 1 5
                       p2 = makePred z 2 6
                       p3 = makePred y 4 8
                       p4 = makePred s 5 9
                       p = liftM4 (&&) p1 p2 p3 p4
                       l = DL.permutations [1..9]

如果未找到匹配项，则

DL.find会返回Nothing，对于匹配列表Just l，会返回l。 fromMaybe将此值转换为空列表或找到的列表。不过，您可以考虑将find的返回值保留为Maybe [Int]。

如何从列表Haskell列表中返回列表

1 个答案: