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70 lines
1.8 KiB
Python
70 lines
1.8 KiB
Python
#!/usr/bin/env python
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# coding=utf-8
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###########################################################################
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# Leetcode 20 有效的括号
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#
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# 给定一个只包括 '(',')','{','}','[',']' 的字符串,判断字符串是否有效。
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# 有效字符串需满足:
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# 1. 左括号必须用相同类型的右括号闭合。
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# 2. 左括号必须以正确的顺序闭合。
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# 注意空字符串可被认为是有效字符串。
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#
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# 示例 1:
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# 输入: "()"
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# 输出: true
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#
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# 示例 2:
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# 输入: "()[]{}"
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# 输出: true
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#
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# 示例 3:
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# 输入: "(]"
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# 输出: false
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#
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# 示例 4:
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# 输入: "([)]"
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# 输出: false
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#
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# 示例 5:
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# 输入: "{[]}"
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# 输出: true
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###########################################################################
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class Solution:
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def isValid(self, s):
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"""
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(knowledge)
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思路:
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1. 用栈,遇到左括号就推入栈;
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2. 遇到右括号就比对栈顶的括号和其是否匹配;
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- 不匹配则返回False
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- 匹配则将栈顶弹出
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- 如果此时栈为空,则返回False
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3. 循环2直至遍历完成,最后,如果栈不为空则返回False,否则返回True
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"""
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# 这边用list来模拟栈
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stack = []
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# 一个字典,用来记录括号映射
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dict = {'(': ')', '{': '}', '[': ']'}
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for c in s:
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if c in dict:
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stack.append(c)
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elif len(stack) == 0 or dict[stack.pop()] != c:
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return False
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return len(stack) == 0
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if __name__ == '__main__':
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solution = Solution()
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print(solution.isValid("()"), "= True")
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print(solution.isValid("()[]{}"), "= True")
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print(solution.isValid("(]"), "= False")
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print(solution.isValid("([)]"), "= False")
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print(solution.isValid("{()}"), "= True")
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