LeetCode题解(0227)：支持四则运算的基础计算器(Python)

题目：​​原题链接​​（中等）

标签：字符串、栈

LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。

解法一（标准逆波兰表达式求值）：

class Solution:
    def calculate(self, s: str) -> int:
        mark_level = {"+": 0, "-": 0, "*": 1, "/": 1}  # 定义符号优先级

        # 合并多位数字
        group = []  # 合并数字组
        for ch in s:
            if ch.isdigit():
                if group and group[-1].isdigit():
                    group[-1] += ch
                else:
                    group.append(ch)
            else:
                group.append(ch)

        # 将中缀表达式转换为后缀表达式
        expression = []  # 表达式
        stack = []  # 符号栈
        for ch in group:
            if ch.isdigit():
                expression.append(ch)
            elif ch in mark_level:
                while stack and mark_level[stack[-1]] >= mark_level[ch]:
                    expression.append(stack.pop())
                stack.append(ch)
        while stack:
            expression.append(stack.pop())

        # 后缀表达式求值
        stack = []  # 数字栈
        for ch in expression:
            if ch.isdigit():
                stack.append(int(ch))
            else:
                b = stack.pop()
                a = stack.pop()
                if ch == "+":
                    stack.append(a + b)
                elif ch == "-":
                    stack.append(a - b)
                elif ch == "*":
                    stack.append(a * b)
                elif ch == "/":
                    stack.append(a // b)

        return stack[0]

解法二（直接用栈求解）：

class Solution:
    def calculate(self, s: str) -> int:
        marks = {"+", "-", "*", "/"}
        stack = []  # 多项式栈
        now_num = ""  # 当前数字
        last_mark = "+"  # 上一个符号
        for ch in s + "+":  # 在结尾添加无意义的运算符，使最后一个数字可以被计算
            if ch.isdigit():
                now_num += ch
            elif ch in marks:
                num = int(now_num)
                if last_mark == "+":
                    stack.append(num)
                elif last_mark == "-":
                    stack.append(-num)
                elif last_mark == "*":
                    stack[-1] *= num
                else:
                    stack[-1] = int(stack[-1] / num)
                now_num = ""
                last_mark = ch

        return sum(stack)