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1- 问题描述
Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.
The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
2- 思路分析[1]
利用一个栈来保存前括号，然后有后括号来时弹出栈顶来判断。
3- Python实现
1 class Solution: 2 # @param {string} s 3 # @return {boolean} 4 def isValid(self, s): 5 if not s: return False 6 l = len(s) 7 if l % 2 == 1: return False # 长度为奇数返回False 8 bracket = {‘(‘:‘)‘, ‘[‘:‘]‘, ‘{‘:‘}‘} 9 stack = [] # 栈保存前括号 10 for i in range(l): 11 if s[i] in bra: 12 stack.append(s[i]) # 前括号则入栈 13 else: 14 try: 15 top = stack.pop() 16 except: 17 return False # 取不出前括号返回False 18 if bracket[top] != s[i]: # 比较是否与出栈前括号匹配 19 return False 20 if not len(stack): return True # 遍历完若栈为空，则完全匹配 21 return False # 栈内还有元素，则不匹配
[1] [LeetCode] Valid Parentheses
Valid Parentheses [LeetCode 20]
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原文地址：http://www.cnblogs.com/freyr/p/4504050.html
