Valid Parentheses


  1. Initialize a stack S: 初始化栈

  2. Process each bracket(括号) of the expression one at a time.

  3. If we encounter an opening bracket, then we check the element on the top of the stack. (遇到左括号则入栈)

  4. If the element at the top of the stack is an openning bracket of the same type, the we pop it off the stack and continue processing. (栈顶元素和外面相匹配,则出栈继续)

  5. Else this implies an invaild expression.

  6. In the end, if we are left with a stack still having elements, then this implies an invaild expression. (栈不空则表达式非法)


def isValid(s):
    stack = []
    mapping = {"]":"[", "}":"{", ")":"("}

    for char in s:
        if char in mapping.keys():
            top_element = stack.pop() if stack else '#'
            if mapping[char] != top_element:
                return False
    return not stack
# s = "()"

Solution of C++


  • 我们遍历字符串 s, 遇到左括号则入栈,遇到右括号 (keys) 则弹出栈顶元素进行比较(在栈非空的前提下)
  • 最终返回值:栈空则合法,等价于 return stack==[]
Last Updated: 12/13/2018, 2:59:53 PM