[leetcode] 20. Valid Parentheses @ python-优快云博客

本文链接：https://blog.youkuaiyun.com/danspace1/article/details/86524924

本文介绍了一种使用字典和堆栈数据结构来验证括号字符串是否有效的方法。该算法确保每种类型的开括号都能被正确闭合，并遵循正确的顺序。通过五个实例演示了算法的有效性。

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原题

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

解法

使用字典保存符号的映射, 使用堆栈保存还没配对的符号, 遍历列表, 如果堆栈中没有正括号与反括号配对或者正括号与反括号不匹配, 返回假, 最后检查堆栈的长度是否为0.
Time: O(n)
Space: O(1)

代码

class Solution:
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        # base case
        if len(s) == 0: return True
        d = {')':'(', '}': '{', ']':'['}
        stack = []
        for char in s:
            if char not in d:
                stack.append(char)
            else:
                if len(stack) == 0 or d[char] != stack.pop():
                    return False
                
        return len(stack) == 0