224. Basic Calculator

时不时写点设计题

题干：

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

解释：

给一个字符串表达式，要求计算他的返回值

思考：

首先,使用3个标志位，sign代表接下来的表达式是正还是负，每次运算结束之后置为1，res代表当前表达式的值，num代表当前数字的值，每次运算之后要清零。遇到括号就初始化，并且把表达式的正负保存下来，压入栈中。遇到右括号再取出来即可。 代码非常清晰简介。89.6

答案：

from collections import deque
class Solution(object):
    def calculate(self, s:str):
        """
        :type s: str
        :rtype: int
        """
        res = 0
        sign = 1
        num = 0
        stack =[]
        
        for c in s:
            if c.isdigit():
                #解决了连续数字的问题
                num = num*10+int(c)
            elif c == '+' or c == '-':
                res = res+num*sign
                sign =1 if c =='+' else -1
                num = 0
            elif c =='(':
                stack.append(res)
                stack.append(sign)
                sign =1 
                res =0
            elif c ==')':
                res = res + sign*num
                old_sign = stack.pop()
                old_res = stack.pop()
                res =old_res+old_sign*res
                sign =1
                num =0
        res = res+sign*num
        return res

答案补充：