Question

我需要使用堆栈检查字符串是否具有平衡的括号，但我的测试用例之一是错误的，因为括号甚至不相等，但返回的是True。

class Stack:
    def __init__(self):
        self.list = []

    def is_empty(self):
        if len(self.list) == 0:
            return True
        else:
            return False
    def push(self, item):
        self.list.append(item)
    def pop(self):
        return self.list.pop()
    def peek(self):
        return self.list[-1]

def balanced_brackets(text):
    s = Stack()
    opening = '(<'
    closing = ')>'
    mapping = dict(zip(opening, closing))

    for letter in text:
        if letter in opening:
            s.push(mapping[letter])
        elif letter in closing:
            if not s or letter != s.pop():
                return False
    return True

'''

print(balanced_brackets('(<x>)(())()'))#this test case is correct
print(balanced_brackets('(((((((xyz))))))'))#but this is wrong

'''

Answer 1

您可以使用collections.Counter对象只计算出括号的数量，然后比较这些数字以确定余额：

from collections import Counter

cnt = Counter('(((((((xyz))))))')
if cnt['('] != cnt[')']:
    print('unbalanced')

我认为您知道了并且可以在此基础上进行构想。

Answer 2

您必须在for循环结束时检查堆栈：

def balanced_brackets(text):
    s = Stack()
    opening = '(<'
    closing = ')>'
    mapping = dict(zip(opening, closing))

    for letter in text:
        if letter in opening:
            s.push(mapping[letter])
        elif letter in closing:
            if s.is_empty() or letter != s.pop():
                return False
    if not s.is_empty():
        return False
    return True

输出：

print(balanced_brackets('(<x>)(())()'))
print(balanced_brackets('(((((((xyz))))))'))

True
False

如何检查字符串中的方括号是否平衡？

2 个答案: