-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy path20.有效的括号.py
73 lines (70 loc) · 1.25 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
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
#
# @lc app=leetcode.cn id=20 lang=python3
#
# [20] 有效的括号
#
# https://leetcode-cn.com/problems/valid-parentheses/description/
#
# algorithms
# Easy (40.92%)
# Likes: 1402
# Dislikes: 0
# Total Accepted: 210.9K
# Total Submissions: 515.3K
# Testcase Example: '"()"'
#
# 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
#
#
# @lc code=start
class Solution:
def isValid(self, s: str) -> bool:
stack = []
match = {'(':')', '{': '}', '[': ']'}
while len(s) > 0:
w = s[0]
s = s[1:]
if w == '(' or w == '{' or w =="[":
stack.append(w)
elif len(stack) == 0 or match[stack.pop()] != w:
return False
return len(stack) == 0
# @lc code=end