LeetCode 224. 基本计算器

给你一个字符串表达式 s ，请你实现一个基本计算器来计算并返回它的值。
注意:不允许使用任何将字符串作为数学表达式计算的内置函数，比如 eval() 。
示例 1：
输入：s = “1 + 1”
输出：2
示例 2：
输入：s = " 2-1 + 2 "
输出：3
示例 3：
输入：s = “(1+(4+5+2)-3)+(6+8)”
输出：23
提示：
1 <= s.length <= 3 * 105
s 由数字、‘+’、‘-’、‘(’、‘)’、和 ’ ’ 组成
s 表示一个有效的表达式
‘+’ 不能用作一元运算(例如， “+1” 和 “+(2 + 3)” 无效)
‘-’ 可以用作一元运算(即 “-1” 和 “-(2 + 3)” 是有效的)
输入中不存在两个连续的操作符
每个数字和运行的计算将适合于一个有符号的 32位 整数

我的解法

class Solution:def calculate(self, s: str) -> int:stack, num, sign = [], None, 1for c in s:# 空白就跳过if c == ' ':continue# 数字就暂存if c in "1234567890":if num is None:num = 0num = num * 10 + int(c)continue# 非数字首先看看是否可以入栈数字if num is not None:stack.append(sign * num)num, sign = None, 1if c == "(":if sign == -1:stack.append("-")sign = 1stack.append(c)elif c in ("+", "-"):if not stack or stack[-1] in ("+", "-", "(", ")"):sign = 2 * (c == "+") - 1else:stack.append(c)else:# c == ")"_num = 0while True:a = stack.pop()if a == "(":stack.append(_num)breakelse:if stack[-1] in ("+", "-"):_num += a * (2 * (stack.pop() == "+") - 1)else:_num += aif num is not None:stack.append(sign * num)num, sign =0, 1for c in stack:if c in ("+", "-"):sign = 2 * (c == "+") - 1else:num += sign * creturn num

参考大佬的进阶解法写一写

from collections import deque
from operator import add, sub, mul, truediv, mod, powclass Solution:def calculate(self, s: str) -> int:# 预处理字符串# 1. 去除空格# 2. 处理单目运算符 +-_s = []for idx, c in enumerate(s):if c == ' ':continueif c in ("+", "-") and (not _s or _s[-1] == "("):_s.append('0')_s.append(c)s = "".join(_s)priority = {"+": 1, "-": 1, "*": 2, "/": 2, "%": 2, "^": 3}operation = {"+": add, "-": sub, "*":mul, "/": lambda a, b: int(truediv(a, b)), "%": mod, "^": pow}nums = deque([])  # 操作数栈ops = deque([]) # 操作符栈i, n = 0, len(s)while i<n:c = s[i]# 解析操作数if c.isdigit():j, _num = i, 0while j < n and s[j].isdigit():_num = _num * 10 + int(s[j])j += 1nums.append(_num)i = j - 1elif c == "(":ops.append(c)elif c in operation.keys():while ops and ops[-1] != "(" and priority[ops[-1]] >= priority[c]:b, a, o = nums.pop(), nums.pop(), ops.pop()nums.append(operation[o](a, b))ops.append(c)else:  # c == ")"while ops and ops[-1] != "(":b, a, o = nums.pop(), nums.pop(), ops.pop()nums.append(operation[o](a, b))if ops and ops[-1] == "(":ops.pop()i += 1while ops and ops[-1] != "(":b, a, o = nums.pop(), nums.pop(), ops.pop()nums.append(operation[o](a, b))return nums.pop()print(Solution().calculate("-1 + (-1+(4+5+2)-3)+(6+8)"))  # '0-1+(0-1+(4+5+2)-3)+(6+8)'