LeetCode: Basic Calculator III

Posted on January 26, 2018July 26, 2020 by braindenny

Basic Calculator III

Similar Problems:

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

The expression string contains only non-negative integers, +, -, *, / operators , open ( and closing parentheses ) and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-2147483648, 2147483647].

Some examples:

"1 + 1" = 2
" 6-4 / 2 " = 4
"2*(5+5*2)/3+(6/2+8)" = 21
"(2+6* 3+5- (3*14/7+2)*5)+3"=-12

Note: Do not use the eval built-in library function.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/basic-calculator-iii
## Basic Ideas: +- yield to */
##
##           When we found one operator as +/, look for the next operator
##           When we found one operator as */, we solve it immediately
## Complexity: Time O(n), Space O(n)
class Solution(object):
    def calculate(self, s):
        """
        :type s: str
        :rtype: int
        """
        stack = []
        s = s.replace(' ', '')
        i, length = 0, len(s)
        while i<length:
            # print stack
            ch = s[i]
            if ch == '(':
                stack.append(ch)
            elif ch == ')':
                str_temp = stack.pop()
                stack.pop() # (
                str_temp = str(self.calculate_no_parenthese(str_temp))
                if len(stack) != 0 and (stack[-1] != '('):
                    stack[-1] = "%s%s" % (stack[-1], str_temp)
                else:
                    stack.append(str_temp)
            else:
                str_temp = ''
                while i<length and s[i] not in '()':
                    str_temp = '%s%s' % (str_temp, s[i])
                    i += 1
                if len(stack) != 0 and (stack[-1] != '('):
                    stack[-1] = "%s%s" % (stack[-1], str_temp)
                else:
                    stack.append(str_temp)
                continue
            i += 1
        return self.calculate_no_parenthese(stack[0])

    def calculate_no_parenthese(self, s):
        """
        :type s: str
        :rtype: int
        """
        """
        :type s: str
        :rtype: int
        """
        s = s.replace(' ', '')
        s = s.replace('--', '+')
        i = 0
        length = len(s)
        # solve */
        stack = []
        while i<length:
            if s[i].isdigit():
                # get the num
                num_str = ''
                while i<length and s[i].isdigit():
                    num_str = "%s%s" % (num_str, s[i])
                    i += 1
                stack.append(num_str)
            elif s[i] in '*/':
                num1 = int(stack.pop())
                num_str = ''
                op = s[i]
                # find the next number
                i += 1
                while i<length and s[i].isdigit():
                    num_str = "%s%s" % (num_str, s[i])
                    i += 1
                num2 = int(num_str)
                if op == '*':
                    num1 = num1*num2
                else:
                    num1 = num1/num2
                stack.append(str(num1))
            else:
                # +-
                stack.append(s[i])
                i += 1

        # solve +-
        res, i = 0, 0
        while i<len(stack):
            element = stack[i]
            if element in '+-':
                num2 = stack[i+1]
                i = i+2
                if element == '+':
                    res += int(num2)
                else:
                    res -= int(num2)
            else:
                res += int(element)
                i += 1
        return res

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