Basic calculator II

Time: O(N); Space: O(N); medium

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only: * non-negative integers, * +, -, , / operators and empty spaces.

The integer division should truncate toward zero.

Example 1:

Input: s =“3 + 2 * 2”

Output: 7

Example 2:

Input: s = ” 3/2 “

Output: 1

Example 3:

Input: s = ” 3+5 / 2 “

Output: 5

Notes:

  • You may assume that the given expression is always valid.

  • Do not use the eval built-in library function.

[1]:

class Solution1(object):

    def compute(self, operands, operators):
        left, right = operands.pop(), operands.pop()
        op = operators.pop()
        if op == '+':
            operands.append(left + right)
        elif op == '-':
            operands.append(left - right)
        elif op == '*':
            operands.append(left * right)
        elif op == '/':
            operands.append(left // right)

    def calculate(self, s: str) -> int:
        """
        :type s: str
        :rtype: int
        """
        operands, operators = [], []
        operand = ""
        for i in reversed(range(len(s))):

            if s[i].isdigit():
                operand += s[i]
                if i == 0  or not s[i-1].isdigit():
                    operands.append(int(operand[::-1]))
                    operand = ""

            elif s[i] == ')' or s[i] == '*' or s[i] == '/':
                operators.append(s[i])

            elif s[i] == '+' or s[i] == '-':
                while operators and \
                      (operators[-1] == '*' or operators[-1] == '/'):
                    self.compute(operands, operators)
                operators.append(s[i])

            elif s[i] == '(':
                while operators[-1] != ')':
                    self.compute(operands, operators)
                operators.pop()

        while operators:
            self.compute(operands, operators)

        return operands[-1]

[2]:

sol = Solution1()
s = "3 + 2 * 2"
assert sol.calculate(s) == 7
s = " 3/2 "
assert sol.calculate(s) == 1
s = " 3+5 / 2 "
assert sol.calculate(s) == 5