20. Valid Parentheses

最新推荐文章于 2024-02-05 23:11:57 发布

emmmxuan

最新推荐文章于 2024-02-05 23:11:57 发布

阅读量133

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签： LeetCode

本文链接：https://blog.youkuaiyun.com/qq_36341799/article/details/83349432

leetcode 专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种使用堆栈数据结构来验证字符串中括号是否正确配对的方法。通过遍历输入字符串，对于每种类型的开括号，将其压入堆栈；对于闭括号，则检查堆栈顶的开括号是否与其匹配，若不匹配则返回错误。最终，如果堆栈为空，说明所有括号都已正确闭合。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

"""
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true
Example 2:

Input: "()[]{}"
Output: true
Example 3:

Input: "(]"
Output: false
Example 4:

Input: "([)]"
Output: false
Example 5:

Input: "{[]}"
Output: true
"""
class Solution(object):
    def isValid(self,s):
        """
        :type s: str
        :rtype: bool
        """
        stack = []
        dict = {")": "(", "]": "[", "}": "{"}
        for i in s:
            if i in dict.values():
                stack.append(i)
            elif i in dict.keys():
                if stack == [] or dict[i] != stack.pop():
                    return False

        return stack == []