Haskell 对字符串中的字符进行递归

Posted 2023-03-16

【中文标题】Haskell 对字符串中的字符进行递归

【英文标题】：Haskell make recursion for Chars in String

【发布时间】：2021-12-31 10:44:03

【问题描述】：

我想在 Haskell 中创建游戏 Magic 15 Puzzle 我有函数 set :: [[Char]] -> Char -> [[Char]] 。 它用 [[Char]] 中的空格切换 Char。

*Main> pp puzzle2
AC DE
FBHIJ
KGLNO
PQMRS
UVWXT
*Main> pp (set puzzle2 'C')
A CDE
FBHIJ
KGLNO
PQMRS
UVWXT
*Main> 

现在我想像这样对[Char]（或String）进行递归（为以前的set x 做set xs）

puzzle :: Result -> [Char] -> Result
puzzle gameboard (x:xs) =  set (set (x:xs) x) xs 

但是编译说是错误的：

 Couldn't match expected type ‘Char’ with actual type ‘[Char]’

我期望这个输出：

*Main> pp(puzzle puzzle2 "CB")                    
ABCDE
F HIJ
KGLNO
PQMRS
UVWXT
      

我能做些什么来解决这个问题？非常感谢您的回答！

完整代码：

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 

【参考方案1】：

把最后一个函数改成

puzzle :: Result -> [Char] -> Result
puzzle g [] = g
puzzle g (x:xs) =  puzzle (set g x) xs 

【讨论】：

非常感谢，它有效，puzzle g [] = g 是什么意思？

这意味着对于空字符串它不会改变，是吗？

是的，就是递归结束的条件。