772. Basic Calculator III.py

class Solution:
    """
    "2*(5+5*2)/3+(6/2+8)"
               i
    """
    def calculate(self, s: str) -> int:
        def cal_update(op, num, stack):
            if op == "+":
                stack.append(num)
            elif op == "-":
                stack.append(-num)
            elif op == "*":
                stack.append(stack.pop()*num)
            elif op == "/":
                stack.append(int(stack.pop()/num))
        s = s+"+"
        n = len(s)
        i = 0
        num = 0
        stack = []
        op = "+"
        while i < n:
            ch = s[i]
            if ch.isdigit():
                num*=10
                num+=ord(ch)-ord("0")
            elif ch in set("+-*/"):
                cal_update(op, num, stack)
                num = 0
                op = ch
            elif ch == "(":
                stack.append(op)
                num=0
                op="+"
            elif ch == ")":
                cal_update(op, num, stack)
                res_in_paren = 0
                while stack[-1] not in set("+-*/"):
                    res_in_paren+=stack.pop()
                op = stack.pop() # 左括號旁邊的符號
                cal_update(op, res_in_paren, stack)
                #num = 0 # 右括號後不會直接遇到數子肯定是+-*/所以不需要把num歸0也不影響
                #不要忘記更新當前op, 當單前op是右括號, 下次再做cal_update就不會放入stack任何res
                op = ch
            i+=1
        return sum(stack)