Leetcode: Basic Calculator

Posted on February 20, 2018September 10, 2019 by braindenny

Basic Calculator

Similar Problems:

Implement a stack 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 .

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23
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.

## Blog link: https://code.dennyzhang.com/basic-calculator
## Basic Ideas: stack
##        Whenever we get ), we get the result for (...)
##
##        # a - b - c != a - (b-c)
##
## Complexity: Time O(n), Space O(n)
class Solution(object):
    def calculate(self, s):
        """
        :type s: str
        :rtype: int
        """
        s = s.replace(' ', '')
        length = len(s)

        if length == 0: return None

        stack = []
        # we can't rely on iterator
        i = 0
        while i < length:
            if s[i] in ['(', '-', '+']:
                stack.append(s[i])
            elif s[i] == ')':
                # recursive removal
                l = []
                while len(stack) != 0 and stack[-1] != '(':
                    l.insert(0, stack.pop())
                stack.pop()
                stack.append(self.calculateList(l))
            else:
                # digit: get longest token for the number
                num_str = ''
                while i < length:
                    if s[i].isdigit() is False: break
                    num_str = '%s%s' % (num_str, s[i])
                    i += 1
                stack.append(num_str)
                continue
            i = i + 1
        # we might get expression without parentheses: 1+3+2
        return self.calculateList(stack)

    def calculateList(self, l):
        # calculate: 2-3+5-7
        res, i = 0, 0
        while i < len(l):
            element = l[i]
            if element in ['-', '+']:
                num2 = int(l[i+1])
                i += 2
                if element == '-':
                    res -= num2
                else:
                    res += num2
            else:
                res += int(element)
                i += 1
        return res

# s = Solution()
# print s.calculate("(1+(4+5+2)-3)+(6+8)") # 23

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