leetcode_day01

任务一：有效的括号

题目链接：https://leetcode-cn.com/problems/valid-parentheses/

自己的答案：

 class Solution:
     def isValid(self, s):
         s = list(s)
         length = len(s)

         #空字符串被认为是有效字符串
         if length == 0:
             return True

         stringList = ['(', ')', '[', ']', '{', '}']
         for s_element in s:
             if  s_element not in stringList:
                 return False

         counter1 = 0
         counter2 = 0
         counter3 = 0
         for s_ele in s:
             if s_ele == "(":
                 counter1 += 1
             elif s_ele == ")":
                 counter1 -= 1
             elif s_ele == "[":
                 counter2 += 1
             elif s_ele == "]":
                 counter2 -= 1
             elif s_ele == "{":
                 counter3 += 1
             elif s_ele == "}":
                 counter3 -= 1
         if counter1 == 0 and counter2 == 0 and counter3 == 0:
             return True
         else:
             return False
         

官方答案：

 class Solution(object):
     def isValid(self, s):
         """
         :type s: str
         :rtype: bool
         """

         # The stack to keep track of opening brackets.
         stack = []

         # Hash map for keeping track of mappings. This keeps the code very clean.
         # Also makes adding more types of parenthesis easier
         mapping = {")": "(", "}": "{", "]": "["}

         # For every bracket in the expression.
         for char in s:

             # If the character is an closing bracket
             if char in mapping:

                 # Pop the topmost element from the stack, if it is non empty
                 # Otherwise assign a dummy value of '#' to the top_element variable
                 top_element = stack.pop() if stack else '#'

                 # The mapping for the opening bracket in our hash and the top
                 # element of the stack don't match, return False
                 if mapping[char] != top_element:
                     return False
             else:
                 # We have an opening bracket, simply push it onto the stack.
                 stack.append(char)

         # In the end, if the stack is empty, then we have a valid expression.
         # The stack won't be empty for cases like ((()
         return not stack