【Leetcode】计算器

思路

用栈来完成；

• 考虑到运算关系，先乘除后加减；
• 此外，一般计算式首个数字式正数；
• 判断字符是否为数字，str.isdigit()
• 字符转数字：ord(str) - ord(‘0’)
• 遇到加减符，压栈数字；遇到乘除符，先运算再压栈
• 遍历完后，再将栈内数字求和

代码

class Solution:def calculate(self, s: str) -> int:stack = []n = len(s)preSign = '+'num = 0for i in range(n):if s[i] != ' ' and s[i].isdigit():num = num * 10 + ord(s[i]) - ord('0')if i == n - 1 or s[i] in '+-*/':if preSign == '+':stack.append(num)elif preSign == '-':stack.append(-num)elif preSign == '*':stack.append(stack.pop() * num)else:stack.append(stack.pop // num)preSign = s[i]num = 0return sum(stack)