圆括号配对（{}[]（<>）问题

Question

我希望能够将字符串中的所有括号配对，如果它们没有配对，那么它们将得到索引号和False。它似乎在一遍又一遍地重复一些值，即cl==pop[1]。我试过看问题在哪里，但不管我怎么努力，都看不出来。所以我问是否有人能帮我找到错误，甚至改进我的代码；）

def check_parentheses(string):
    pending = 0
    brackets = []
    '''Checks if parens are paired, otherwise they are bad.'''
    parenstack = collections.deque()
    for ch in string:
        if ch in lrmap:
            try:
                cl = string.index(ch, pending)
                pending = cl + 1

            except:
                cl = False

        if ch in lparens:
            parenstack.append([ch, cl])
            print parenstack

        elif ch in rparens:
            try:
                pop = parenstack.pop()

                if lrmap[pop[0]] != ch:
                    print 'wrong type of parenthesis popped from stack',\
                    pop[0], ch, pop[1], cl

                    brackets.append([pop[1], False])
                    brackets.append([cl, False])
                else:
                    brackets.append([pop[1], cl])

            except IndexError:
                print 'no opening parenthesis left in stack'
                brackets.append([cl, False])

    # if we are not out of opening parentheses, we have a mismatch
    for p in parenstack:
        brackets.append([p[1],False])
    return brackets

Answer 1

iparens = iter('(){}[]<>')
parens = dict(zip(iparens, iparens))
closing = parens.values()

def balanced(astr):
    stack = []
    for c in astr:
        d = parens.get(c, None)
        if d:
            stack.append(d)
        elif c in closing:
            if not stack or c != stack.pop():
                return False
    return not stack

示例：

>>> balanced('[1<2>(3)]')
True
>>> balanced('[1<2(>3)]')
False

Answer 2

您可以将我的代码修改为类似的问题：

def Evaluate(str):
  stack = []
  pushChars, popChars = "<({[", ">)}]"
  for c in str :
    if c in pushChars :
      stack.append(c)
    elif c in popChars :
      if not len(stack) :
        return False
      else :
        stackTop = stack.pop()
        balancingBracket = pushChars[popChars.index(c)]
        if stackTop != balancingBracket :
          return False
    else :
      return False
  return not len(stack)

Answer 3

BRACES = { '(': ')', '[': ']', '{': '}' }

def group_check(s):
    stack = []
    for b in s:
        c = BRACES.get(b)
        if c:
            stack.append(c)
        elif not stack or stack.pop() != b:
            return False
    return not stack