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给定一个只包括 '('
,')'
,'{'
,'}'
,'['
,']'
的字符串,判断字符串是否有效。
有效字符串需满足:
- 左括号必须用相同类型的右括号闭合。
- 左括号必须以正确的顺序闭合。
注意空字符串可被认为是有效字符串。
示例 1:
输入: "()"
输出: true
示例 2:
输入: "()[]{}"
输出: true
示例 3:
输入: "(]"
输出: false
示例 4:
输入: "([)]"
输出: false
示例 5:
输入: "{[]}"
输出: true
class Solution(object):
def isValid(self, s):
"""
:type s: str
:rtype: bool
"""
lst = []
for st in s:
if st == "(" or st == "["or st == "{":
lst.append(st)
else:
if (st == ")" and lst[-1] =="(")\
or (st == "]" and lst[-1] =="[")\
or (st == "}" and lst[-1] =="{"):
lst.pop()
else:
return bool(0)
###下面两句可换成 return len(lst)==0
if len(lst) == 0:
return bool(1)
else:
return bool(0)
class Solution(object):
def isValid(self, s):
"""
:type s: str
:rtype: bool
"""
#### 键值对正写很麻烦,所以采用下面到写 字典只能通过key找value########
a = {')':'(', ']':'[', '}':'{'}
#### l为空l[-1]会出现indexerror,len(l[None])为 ####
l = [None]
for i in s:
if i in a and a[i] == l[-1]:
l.pop()
else:
l.append(i)
return len(l)==1
class Solution(object):
def isValid(self, s):
"""
:type s: str
:rtype: bool
"""
dic = {')':'(', ']':'[', '}':'{'}
lst = []
for st in s:
if st in dic and len(lst) != 0 and lst[-1]==dic[st]:
lst.pop()
elif st in dic and len(lst) != 0 and lst[-1]!=dic[st]:
return bool(0)
else:
lst.append(st)
return len(lst)==0