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LeetCode 227 Basic Calculator II - Medium

227. Basic Calculator II -- Medium

·1 min read·#LeetCode#Medium#Python

227. Basic Calculator II — Medium

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Problem

  1. Basic Calculator II -- Medium

Given a string s which represents an expression, evaluate this expression and return its value.

The integer division should truncate toward zero.

Example 1:

Input: s = "3+2*2" Output: 7 Example 2:

Input: s = " 3/2 " Output: 1 Example 3:

Input: s = " 3+5 / 2 " Output: 5

Constraints:

1 <= s.length <= 3 * 105 s consists of integers and operators ('+', '-', '*', '/') separated by some number of spaces. s represents a valid expression. All the integers in the expression are non-negative integers in the range [0, 231 - 1]. The answer is guaranteed to fit in a 32-bit integer.

Solution

class Solution:
    def calculate(self, s: str) -> int:
        #通过栈来实现运算，按顺序读取字符串，
        #将第一个数入栈。之后遇到+，将下一个数num入栈；遇到-，则将-num入栈；
        #遇到乘或除，先将上一个数出栈，与当前数进行运算后，再将结果入栈。
        #读取完整个字符串后，将栈中所有的数相加即运算结果。
        if not s:
            return "0"
        stack, num, sign = [], 0, "+"
        for i in range(len(s)):
            if s[i].isdigit():
                num = num*10+ord(s[i])-ord("0")
            if (not s[i].isdigit() and not s[i].isspace()) or i == len(s)-1:
                if sign == "-":
                    stack.append(-num)
                elif sign == "+":
                    stack.append(num)
                elif sign == "*":
                    stack.append(stack.pop()*num)
                else:
                    tmp = stack.pop()
                    if tmp//num < 0 and tmp%num != 0:
                        stack.append(tmp//num+1)
                    else:
                        stack.append(tmp//num)
                sign = s[i]
                num = 0
        return sum(stack)

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