README.md

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.

Solutions (Python)

1. Solution

class Solution:
    def calculate(self, s: str) -> int:
        s = s.replace(' ', '')
        negstack = [False]
        stack = []
        ret = 0

        for i in range(len(s)):
            if s[i] == '(':
                if i == 0 or s[i - 1] != '-':
                    negstack.append(negstack[-1])
                else:
                    negstack.append(not negstack[-1])
            elif s[i] == ')':
                negstack.pop()
            elif s[i] == '+':
                stack.append('-' if negstack[-1] else '+')
            elif s[i] == '-':
                if stack == [] or isinstance(stack[-1], str):
                    stack.append(0)
                stack.append('+' if negstack[-1] else '-')
            else:
                if stack == [] or isinstance(stack[-1], str):
                    stack.append(0)
                stack[-1] = stack[-1] * 10 + int(s[i])

        ret = stack[0]

        for i in range(2, len(stack), 2):
            ret += stack[i] if stack[i - 1] == '+' else -stack[i]

        return ret