标签：log port utf-8 from 是否有效 balance find 共享 and
题目：
字符串中有括号”()[]{}”，设计算法，判断该字符串是否有效
括号必须以正确的顺序配对，如：“()”、“()[]”是有效的，但“([)]”无效
解法一：
# coding=utf-8
from pythonds.basic.stack import Stack # 栈可以不用此包，入栈append，出栈pop
def parChecker(symbolString):
s = Stack()
balanced = True
symbolList = list(symbolString)
index = 0
while balanced == True and index < len(symbolString):
if symbolString[index] == "(":
s.push(symbolString[index])
else: # symbolString=")"
if s.isEmpty():
balanced = False
else:
s.pop()
index += 1
if s.isEmpty():
balanced = True
else:
balanced = False
# for i in symbolList:
# if i == ‘(‘:
# s.push(i)
# else: # i == ‘)‘
# if s.isEmpty():
# balanced = False
# else:
# s.pop()
# if s.isEmpty():
# balanced = True
# else:
# balanced = False
return balanced
if __name__ == "__main__":
symbolString = ‘(())()()()((()()(()())))(‘
print(parChecker(symbolString))
解法二（更好）：（来自https://blog.csdn.net/weixin_42018258/article/details/80579081）
def match_parentheses(s):
# 把一个list当做栈使用
ls = []
parentheses = "()[]{}"
for i in range(len(s)):
si = s[i]
# 如果不是括号则继续
if parentheses.find(si) == -1:
continue
# 左括号入栈
if si == ‘(‘ or si == ‘[‘ or si == ‘{‘:
ls.append(si)
continue
if len(ls) == 0:
return False
# 出栈比较是否匹配
p = ls.pop()
if (p == ‘(‘ and si == ‘)‘) or (p == ‘[‘ and si == ‘]‘) or (p == ‘{‘ and si == ‘}‘):
continue
else:
return False
if len(ls) > 0:
return False
return True
if __name__ == ‘__main__‘:
s = "{abc}{de}(f)[(g)"
result = match_parentheses(s)
print(s, result)
s = "0{abc}{de}(f)[(g)]9"
result = match_parentheses(s)
print(s, result)
标签：log port utf-8 from 是否有效 balance find 共享 and
原文地址：https://www.cnblogs.com/sddai/p/13931209.html
