验证括号匹配的Haskell函数

0lvr5msh  于 2023-10-19  发布在  其他
关注(0)|答案(7)|浏览(103)

我需要写一个函数par :: String -> Bool来验证一个给定的带括号的字符串是否匹配堆栈模块。
例如:

par "(((()[()])))" = True
par "((]())" = False

下面是我的stack模块实现:

module Stack (Stack,
              push, pop, top,
              empty, isEmpty)
    where

data Stack a = Stk [a]
             deriving (Show)

push :: a -> Stack a -> Stack a
push x (Stk xs) = Stk (x:xs)

pop :: Stack a -> Stack a
pop (Stk (_:xs)) = Stk xs
pop _ = error "Stack.pop: empty stack"

top :: Stack a -> a
top (Stk (x:_)) = x
top _ = error "Stack.top: empty stack"

empty :: Stack a
empty = Stk []

isEmpty :: Stack a -> Bool
isEmpty (Stk [])= True
isEmpty (Stk _) = False

因此,我需要实现一个par函数,它将测试一个括号字符串,并判断其中的括号是否平衡。我怎么能用一个堆栈来实现这一点呢?

eivnm1vs

eivnm1vs1#

module Parens where

import Data.Map (Map)
import qualified Data.Map as Map

matchingParens :: Map Char Char
matchingParens = Map.fromList [
    ('(', ')')
  , ('{', '}')
  , ('[', ']')
  ]

isOpening :: Char -> Bool
isOpening c = maybe False (const True) $ Map.lookup c matchingParens

type Stack a = [a]

balanced :: String -> Bool
balanced = balanced' []

balanced' :: Stack Char -> String -> Bool
balanced' [] ""     = True
balanced' _  ""     = False
balanced' [] (c:cs) = balanced' [c] cs
balanced' (o:os) (c:cs)
  | isOpening c = balanced' (c:o:os) cs
  | otherwise   = case Map.lookup o matchingParens of
      Nothing -> False
      Just closing -> if closing == c
        then balanced' os cs
        else False
8fsztsew

8fsztsew2#

答案如下:

parent' :: String -> Stack Char -> Bool
parent' [] stk = isEmpty stk
parent' (c:str) stk
        | (c == '(') = parent' str (push c stk)
        | (c == ')') = if isEmpty stk then False
                       else if top stk == '(' then parent' str (pop stk)
                       else False


parent :: String -> Bool
parent [] = True
parent str = parent' str empty
nimxete2

nimxete23#

import Data.Maybe
import Control.Monad

parse :: String -> Maybe String
parse xs@(')':_) = return xs
parse xs@(']':_) = return xs
parse ('(':xs) = do
  ')':ys <- parse xs
  parse ys
parse ('[':xs) = do
  ']':ys <- parse xs
  parse ys
parse (_:xs) = parse xs
parse [] = return []

paren :: String -> Bool
paren xs = isJust $ do
  ys <- parse xs
  guard (null ys)
3j86kqsm

3j86kqsm4#

我是一个 haskell 新手。这是我的尝试,绝对不优雅,但想尝试不同的方法

data Stack a = Stk [a]
         deriving (Show)

push :: a -> Stack a -> Stack a
push x (Stk xs) = Stk (x:xs)

pop :: Stack a -> (Maybe a, Stack a)
pop (Stk []) = (Nothing, Stk [])
pop (Stk (x:xs)) = (Just x, Stk xs)

top :: Stack a -> Maybe a
top (Stk (x:_)) = Just x
top _ = Nothing

empty :: Stack a
empty = Stk []

isEmpty :: Stack a -> Bool
isEmpty (Stk [])= True
isEmpty (Stk _) = False 

par :: String -> Maybe (Stack Char)
par = foldl check (Just (Stk []))
      where check (Just stk) x
                | x == '(' = Just (push x stk)
                | x == ')' = case pop stk of
                                     (Just '(', newStk) -> Just newStk
                                     _ -> Nothing
            check Nothing x = Nothing

parCheck :: String -> Bool
parCheck xs = case par xs of
                Just stk -> isEmpty stk
                Nothing -> False
zpjtge22

zpjtge225#

parent :: String -> Bool
parent "" = True
parent str = verify str empty

verify :: String -> Stack Char -> Bool
verify [] stk = isEmpty stk
verify (c:str) stk
        | (c == '(') = verify str (push c stk)
        | (c == ')') = if isEmpty stk then False else if top stk == '(' then verify str (pop stk) else False
        | (c == '[') = verify str (push c stk)
        | (c == ']') = if isEmpty stk then False else if top stk == '[' then verify str (pop stk) else False
dgjrabp2

dgjrabp26#

import Data.Char

verifier :: String -> Bool
verifier x = balancer x []
    where
balancer [] stack = null stack
balancer (x:xs) [] = balancer xs [x]
balancer (x:xs) (y:ys)  = if isSpace x then balancer xs (y:ys)
              else if x `elem` "([{" then balancer xs (x:y:ys)
                  else if (x == ')' && y == '(') ||  
                  (x == ']' && y == '[') || 
                  (x == '}' && y == '{')  then balancer xs ys 
                else False
xxls0lw8

xxls0lw87#

isOpenBracket :: Char -> Bool
isOpenBracket x = case x of 
    '{' -> True
    '[' -> True
    '(' -> True
    otherwise -> False  

isCloseBracket :: Char -> Bool
isCloseBracket x = case x of 
    '}' -> True
    ']' -> True
    ')' -> True
    otherwise -> False 

bracketLookup :: Char -> Char
bracketLookup x = case x of 
    '{' -> '}'
    '[' -> ']'
    '(' -> ')'

checkBrackets :: [Char] -> [Char] -> Bool 
checkBrackets [] []             = True
checkBrackets [] (x:xs)         = (isOpenBracket x) && (checkBrackets (x:[]) (xs))
checkBrackets (s:ss) (x:xs) 
    | (isOpenBracket x)         = (checkBrackets (x:s:ss) (xs))
    | (isCloseBracket x)        = (x == bracketLookup s) && checkBrackets (ss) (xs)
    -- case where x is not a a bracket (invalid char), just skip it
    | otherwise                 = checkBrackets (s:ss) (xs) 
checkBrackets (x:xs) [] = False

bracketParity x = checkBrackets "" x

我的尝试,只是使用一个列表作为一个堆栈,并重新循环沿着输入。忽略非括号字符。
输出:

*Main> bracketParity "{()([a])}"
True
*Main> bracketParity "{()([aaaa])}"
True
*Main> bracketParity "{(a)([aaaa]a)}"
True
*Main> bracketParity "{()([])}"
True
*Main> bracketParity "[][][]"
True
*Main> bracketParity "[][][]]"
False

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