Question

我需要编写一个函数par :: String -> Bool来验证带有括号的给定字符串是否与堆栈模块匹配。

例如：

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

这是我的堆栈模块实现：

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

所以我需要实现一个par函数来测试一串括号，并说明它中的括号是否平衡。我怎么能用堆叠做到这一点？

Answer 1

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

Answer 2

以下是答案：

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

Answer 3

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)

Answer 4

我是一名哈斯克新手。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

Answer 5

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

Answer 6

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

Haskell函数验证括号匹配

6 个答案: