leetcode227 Basic Calculator II

"""
Implement a basic calculator to evaluate a simple expression string.
The expression string contains only non-negative integers, +, -, *, / 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
"""
"""
这道题因为有空格字符，连续数字字符使得考虑情况有些复杂
为了保存好符号，用了一个sign变量，初始化为'+'，保证其正常入栈
"""
class Solution:
    def calculate(self, s: str) -> int:
        if not s:
            return '0'
        stack = []
        sign = '+'  # !!!用一个sign实现将符号后置使用
        num = 0
        for i in range(len(s)):
            if s[i].isdigit():
                num = num * 10 + int(s[i])  # ord(s[i]) - ord('0')
            if not s[i].isdigit() and s[i] != ' ' or i == len(s) - 1:
                if sign == '-':  # 注意这里是sign
                    stack.append(-num)
                elif sign == '+':
                    stack.append(num)
                elif sign == '*':
                    stack.append(stack.pop() * num)
                else:  # 因为python里向下取整,-1.5取整为-2
                    temp = stack.pop()
                    if temp // num < 0 and temp % num != 0:
                        stack.append(temp // num + 1)
                    else:
                        stack.append(temp // num)
                sign = s[i]
                num = 0
        res = 0
        for j in range(len(stack)):
            res += stack[j]
        return res
if __name__ == '__main__':
    ans = Solution()
    s = "14-3/2"
    ans.calculate(s)