-
Notifications
You must be signed in to change notification settings - Fork 0
/
20.有效的括号.py
51 lines (40 loc) · 1.65 KB
/
20.有效的括号.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
class Solution:
def isValid(self, s):
# 核心思想
# 直接判断字符串是否有连续的括号,有则替换为空
# 最终判断s是否为空
while '{}' in s or '()' in s or '[]' in s:
s = s.replace('{}', '')
s = s.replace('[]', '')
s = s.replace('()', '')
return s == ''
def isValid2(self, s: str) -> bool:
# 核心思想--栈
# 将s中的元素依次入栈,当循环到的元素与栈顶元素匹配
# 则将栈顶元素出栈,若最终栈为空,则说明每一个括号都有匹配的,返回true
if s == '':
# 若s为空,直接返回true
return True
stack = list()
n = len(s)
stack.append(s[0])
if n%2:
# 若s的长度不是2的倍数则直接返回false
# 因为括号必须成双成对出现
return False
for i in range(1, n):
if stack == []:
# 若stack中元素匹配完成后为空,则下个元素必须入栈
stack.append(s[i])
else:
# 判断是否和栈顶元素配对
if stack[-1] == '(' and s[i] == ')':
stack.pop()
elif stack[-1] == '[' and s[i] == ']':
stack.pop()
elif stack[-1] == '{' and s[i] == '}':
stack.pop()
else:
# 若没不和栈顶元素配对,则入栈
stack.append(s[i])
return stack == []