- categories: Code, Interview Question, leetcode, Hard
- source: https://leetcode.com/problems/basic-calculator
- topics: Stack, String Manipulation
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().
Idea
use - operator as the magnitude for the operand to the right of the operator. Once we start using - as a magnitude for the operands, we just have one operator left which is addition and + is associative
class Solution:
def calculate(self, s: str) -> int:
stack = []
operand = 0
res = 0 # For the on-going result
sign = 1 # 1 means positive, -1 means negative
for ch in s:
if ch.isdigit():
# Forming operand, since it could be more than one digit
operand = (operand * 10) + int(ch)
elif ch == '+':
# Evaluate the expression to the left,
# with result, sign, operand
res += sign * operand
# Save the recently encountered '+' sign
sign = 1
# Reset operand
operand = 0
elif ch == '-':
res += sign * operand
sign = -1
operand = 0
elif ch == '(':
# Push the result and sign on to the stack, for later
# We push the result first, then sign
stack.append(res)
stack.append(sign)
# Reset operand and result, as if new evaluation begins for the new sub-expression
sign = 1
res = 0
elif ch == ')':
# Evaluate the expression to the left
# with result, sign and operand
res += sign * operand
# ')' marks end of expression within a set of parenthesis
# Its result is multiplied with sign on top of stack
# as stack.pop() is the sign before the parenthesis
res *= stack.pop() # stack pop 1, sign
# Then add to the next operand on the top.
# as stack.pop() is the result calculated before this parenthesis
# (operand on stack) + (sign on stack * (result from parenthesis))
res += stack.pop() # stack pop 2, operand
# Reset the operand
operand = 0
return res + sign * operand