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,那么它们会得到索引号和 False。看起来它一遍又一遍地重复一些值,即 cl == pop[1]。我试图找出问题出在哪里,但无论我如何努力,我都看不到它。所以我问是否有人帮助我找到错误,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
您可以将我的代码调整为类似的问题:
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)
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