20. Valid Parentheses

EasyStringStackTimeO(n)SpaceO(n)
(
[
{
}
]
)
stack
empty
reading…
read '('read
Frame 1 / 16
1class Solution:
2 def isValid(self, s: str) -> bool:
3 stack = []
4 pairs = {')': '(', ']': '[', '}': '{'}
5 for char in s:
6 if char in pairs:
7 if stack and stack[-1] == pairs[char]:
8 stack.pop()
9 else:
10 return False
11 else:
12 stack.append(char)
13 return not stack