0224-basic-calculator.rb

# frozen_string_literal: true

# 224. Basic Calculator
# https://leetcode.com/problems/basic-calculator
# Hard

=begin
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().

Example 1:
Input: s = "1 + 1"
Output: 2

Example 2:
Input: s = " 2-1 + 2 "
Output: 3

Example 3:
Input: s = "(1+(4+5+2)-3)+(6+8)"
Output: 23

Constraints:
1 <= s.length <= 3 * 105
s consists of digits, '+', '-', '(', ')', and ' '.
s represents a valid expression.
'+' is not used as a unary operation (i.e., "+1" and "+(2 + 3)" is invalid).
'-' could be used as a unary operation (i.e., "-1" and "-(2 + 3)" is valid).
There will be no two consecutive operators in the input.
Every number and running calculation will fit in a signed 32-bit integer.
=end

# @param {String} s
# @return {Integer}
defcalculate(s)
last=s.size - 1
total=0
sign=1
current=0
is_char=->index{s[index].ord.between?(?0.ord, ?9.ord)}
stack=[]
(0...s.size).eachdo |i|
char=s[i]
if ?+ == char
sign=1
elsif ?- == char
sign= -1
elsif" " == char
elsif ?( == char
stack.push([total,sign])
total=0
sign=1
current=0
elsif ?) == char
prev_total,prev_sign=stack.pop()
current=0
total=prev_total + prev_sign * total
sign=0
elsifis_char[i]
current=current * 10 + char.to_i
iflast == i || ! is_char[i + 1]
total += sign * current
current=0
sign=0
end
end
end
total
end

# **************** #
# TEST #
# **************** #

require"test/unit"
classTest_calculate < Test::Unit::TestCase
deftest_
assert_equal2,calculate("1 + 1")
assert_equal3,calculate(" 2-1 + 2 ")
assert_equal23,calculate("(1+(4+5+2)-3)+(6+8)")
end
end