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
         

 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