Basic Calculator II

Implement a basic calculator to evaluate a simple expression string.

The expression string contains onlynon-negativeintegers,+,-,*,/operators and empty spaces. The integer division should truncate toward zero.

Example 1:

Input: 
"3+2*2"

Output:
 7

Example 2:

Input:
 " 3/2 "

Output:
 1

Example 3:

Input:
 " 3+5 / 2 "

Output:
 5

分析

计算机的题目好像都是op在前，数字带着op走，或者入栈，或者入expression带着走。

注意这里2个陷阱：最后一个数也要加入栈，所以ii == n-1 不是else

/的时候 3/2和-3/2要额外处理。用ceil和floor

class Solution:
    def calculate(self, s):
        """
        :type s: str
        :rtype: int
        """
        stack = []
        op ='+'
        snum =''
        n = len(s)
        res = 0
        for ii,i in enumerate(s):
            if i.isdigit():
                snum += i
            if i in "+-*/" or ii == n-1:
                if op == '+':
                    stack.append(int(snum))
                elif op == '-':
                    stack.append(-int(snum))
                elif op == '*':
                    cur = stack.pop()
                    stack.append(cur*int(snum))
                elif op == '/':
                    cur = stack.pop()/int(snum)
                    stack.append(math.floor(cur)) if cur >= 0 else stack.append(math.ceil(cur))
                snum =''
                op = i

        for i in stack :
            res += i


        return res