我想在 Haskell 中創建游戲 Magic 15 Puzzle 我有函式set :: [[Char]] -> Char -> [[Char]]。它在 [[Char]] 中用空格切換 Char。
*Main> pp puzzle2
AC DE
FBHIJ
KGLNO
PQMRS
UVWXT
*Main> pp (set puzzle2 'C')
A CDE
FBHIJ
KGLNO
PQMRS
UVWXT
*Main>
現在我想像這樣為[Char]（或String）做遞回（set為前一個setx做xs ）
puzzle :: Result -> [Char] -> Result
puzzle gameboard (x:xs) = set (set (x:xs) x) xs
但編譯說這是錯誤：
Couldn't match expected type ‘Char’ with actual type ‘[Char]’
我希望這個輸出：
*Main> pp(puzzle puzzle2 "CB")
ABCDE
F HIJ
KGLNO
PQMRS
UVWXT
我能做些什么來解決這個問題？非常感謝您的回答！
全碼：
import Data.Char
type Result = [String]
pp :: Result -> IO ()
pp x = putStr (concat (map ( "\n") x))
puzzle2 :: [[Char]]
puzzle2 = ["AC DE",
"FBHIJ",
"KGLNO",
"PQMRS",
"UVWXT"]
getCords board x = head ( head [[(row_index, column_index) |(column_index, char) <- zip[1..] row, x == char] |(row_index,row)<- zip [1..]board,x `elem` row])
getRow board c = fst ( getCords board c)
getCol board c = snd ( getCords board c)
check ch1 ch2 board = (getRow board ch2 == getRow board ch1 1 || getRow board ch2 == getRow board ch1 - 1) && (getCol board ch1 == getCol board ch2) || ((getRow board ch1 == getRow board ch2) && (getCol board ch2 == getCol board ch1 1 || getCol board ch2 == getCol board ch1 - 1) )
set gameboard x | check x ' ' gameboard = [[if ch == ' ' then x else if ch == x then ' ' else ch | ch<- line] | line<-gameboard]
| not (check x ' ' gameboard ) = [[ch | ch<- line] | line<-gameboard]
puzzle :: Result -> [Char] -> Result
puzzle gameboard (x:xs) = set (set (x:xs) x) xs
uj5u.com熱心網友回復：
只需將最后一個函式更改為
puzzle :: Result -> [Char] -> Result
puzzle g [] = g
puzzle g (x:xs) = puzzle (set g x) xs
轉載請註明出處，本文鏈接：https://www.uj5u.com/qiye/363867.html