20. 有效的括号

算法

1> 初始化栈 S。
2> 一次处理表达式的每个括号。
3> 如果遇到开括号，我们只需将其推到栈上即可。这意味着我们将稍后处理它，让我们简单地转到前面的 子表达式。
4> 如果我们遇到一个闭括号，那么我们检查栈顶的元素。如果栈顶的元素是一个 相同类型的 左括号，那么我们将它从栈中弹出并继续处理。否则，这意味着表达式无效。
5> 如果到最后我们剩下的栈中仍然有元素，那么这意味着表达式无效。

复杂度分析

时间复杂度：O(n)，因为我们一次只遍历给定的字符串中的一个字符并在栈上进行 O(1) 的推入和弹出操作。
空间复杂度: O(n)，当我们将所有的开括号都推到栈上时以及在最糟糕的情况下，我们最终要把所有括号推到栈上。例如 ((((((((((。

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

       

 top_element = stack.pop() if stack else '#'
 等价于
 if stack:
 	top_element = stack.pop()
 else:
 	top_element = '#'

/* !< NO.1 */
class Solution:
    def isValid(self, s):
        while '()' in s or '[]' in s or '{}' in s:
            s = s.replace('()','').replace('[]','').replace('{}','')
        return s == ''  

 stack = []
        mapping = {'(':')', '[':']', '{':'}'}
        open_par = set(['(', '[', '{'])
        for char in s:
            if char in open_par:
                stack.append(char)
            elif stack and char == mapping[stack[-1]]:
                stack.pop()
            else:
                return False
        return stack == []