Leetcode 224. Basic Calculator

Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.

Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Example 1:Input: s = “1 + 1” Output: 2

Example 2:Input: s = ” 2-1 + 2 ” Output: 3

Example 3:Input: s = “(1+(4+5+2)-3)+(6+8)” Output: 23

Constraints:

• 1 <= s.length <= 3 * 105
• s consists of digits, '+', '-', '(', ')', and ' '.
• s represents a valid expression.
• '+' is not used as a unary operation (i.e., "+1" and "+(2 + 3)" is invalid).
• '-' could be used as a unary operation (i.e., "-1" and "-(2 + 3)" is valid).
• There will be no two consecutive operators in the input.
• Every number and running calculation will fit in a signed 32-bit integer.

My solution is a typical way. However, the performance is not that good. Just FYI.

class Solution:
    def calculate(self, s: str) -> int:
        pattern = r'(\d+|\D)'

        tokens = re.findall(pattern, s)

        stack = []

        for char in tokens:
            if char == ' ':
                pass
            elif char == ')':
                sub_stack = self.get_substack(stack)
                stack.append(self.calculate_subtotal(sub_stack))
            else:
                stack.append(char)
        
        return self.calculate_subtotal(stack)
    
    def get_substack(self, stack):
        sub_stack = []

        while stack and stack[-1] != '(':
            sub_stack.append(stack.pop())

        if stack and stack[-1] == '(':
            stack.pop()

        return sub_stack[::-1]

    def calculate_subtotal(self, stack):
        total = 0
        sign = None

        for char in stack:
            if str(char) == '-' or str(char) == '+':
                sign = char
            else:
                if sign == '-':
                    total -= int(char)
                else:
                    total += int(char)
        
        return total