leetcode - 772. Basic Calculator III

Description

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, ‘+’, ‘-’, ‘*’, ‘/’ operators, and open ‘(’ and closing parentheses ‘)’. 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 [-2^31, 2^31 - 1].

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 = "6-4/2"
Output: 4

Example 3:

Input: s = "2*(5+5*2)/3+(6/2+8)"
Output: 21

Constraints:

1 <= s <= 10^4
s consists of digits, '+', '-', '*', '/', '(', and ')'.
s is a valid expression.

Solution

Consider +, - as positive number and negative number, use recursive to solve parentheses. See 227. 基本计算器 II (+, -, *, /，无括号) for details.

Code

class Solution:
    def calculate(self, s: str) -> int:
        def find_closing_index(index: int) -> int:
            left_cnt = 0
            i = index
            while i < len(s):
                if s[i] == '(':
                    left_cnt += 1
                elif s[i] == ')':
                    left_cnt -= 1
                if left_cnt == 0:
                    break
                i += 1
            return i

        def helper(left: int, right: int) -> int:
            if s[left] == '(' and right == find_closing_index(left):
                return helper(left + 1, right - 1)
            op, sign = 0, '+'
            stack = []
            index = left
            while index <= right:
                if s[index].isdecimal():
                    op = op * 10 + int(s[index])
                    index += 1
                elif s[index] in ('+', '-', '*', '/'):
                    if sign in ('+', '-'):
                        stack.append(op * (1 if sign == '+' else -1))
                    elif sign == '*':
                        stack[-1] *= op
                    elif sign == '/':
                        stack[-1] = int(stack[-1] / op)
                    op = 0
                    sign = s[index]
                    index += 1
                elif s[index] == '(':
                    index_right = find_closing_index(index)
                    op = helper(index, index_right)
                    index = index_right + 1
                else:
                    index += 1
            if sign in ('+', '-'):
                stack.append(op * (1 if sign == '+' else -1))
            elif sign == '*':
                stack[-1] *= op
            elif sign == '/':
                stack[-1] = int(stack[-1] / op)
            return sum(stack)
        return helper(0, len(s) - 1)