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LeetCode--020--括号匹配
阅读量:799 次
发布时间:2023-01-31

本文共 742 字,大约阅读时间需要 2 分钟。

为了判断一个包含不同类型括号的字符串是否有效,我们可以使用栈数据结构。栈可以帮助我们跟踪括号的顺序,确保每个左括号都能被正确闭合。

输入包括六种括号字符:'('、')'、'{'、'}'、'['、']'。有效字符串要求左括号必须以正确的顺序和类型闭合。具体来说:

  • 每次遇到左括号,压入栈。
  • 每次遇到右括号,检查栈是否为空:
    • 如果栈为空,字符串无效。
    • 如果栈不为空,弹出栈顶字符,检查是否对应正确的左括号。
  • 遍历完整个字符串后,栈必须为空才表示闭合顺序正确。
  • 实现代码:

    class Solution:    def isValid(self, s: str) -> bool:        stack = []        for c in s:            if c in '({[':                stack.append(c)            elif c in ')]}':                if not stack:                    return False                top = stack.pop()                if (c == ')' and top != '(') or (c == '}' and top != '{') or (c == ']' and top != '['):                    return False        return len(stack) == 0

    这个方法通过遍历字符串一次,检查每个字符是否正确闭合,从而确保字符串的有效性。逻辑简单且效率高,时间复杂度为 O(n),其中 n 是字符串的长度。

    转载地址:http://vlgyk.baihongyu.com/

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