LeetCode20有效的括号

未经博主同意，禁止瞎JB转载。

LeetCode20有效的括号

https://leetcode-cn.com/problems/valid-parentheses/description/

我的解法：

示例如下：

利用了栈的思想，使用两个列表实现配对，与官方提供的解答是一样的，但是写的就比较麻烦。

class Solution(object):
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        # "(([]){})"
        s = list(s)
        n = len(s)
        if n == 0:
            return True
        if n % 2 == 1:
            return False
        ret = []
        i = 0
        while len(s):
            if s[0] == '(' or s[0] == '[' or s[0] == '{':
                ret.append(s.pop(0))
                if len(ret)>len(s):
                    return False
            else:
                if len(ret) == 0:
                    return False
                while len(ret):
                    if ret[-1] == '(' and s[0] == ')' or ret[-1] == '[' and s[0] == ']' or ret[-1] == '{' and s[0] == '}':
                        ret.pop()
                        s.pop(0)
                    elif s[0] == '(' or s[0] == '[' or s[0] == '{':
                        break
                    else:
                        return False
        if len(ret):
            return False
        return True

 官方的解答：

https://leetcode-cn.com/problems/valid-parentheses/solution/

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