LeetCode Top Interview 150
224. Basic Calculator
Hard
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 * 105sconsists of digits,'+','-','(',')', and' '.srepresents 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.
Solution
class Solution:
def calculate(self, s: str) -> int:
stack = []
result = 0
num = 0
sign = 1
for char in s:
if char.isdigit():
num = num * 10 + int(char)
elif char == '+':
result += sign * num
num = 0
sign = 1
elif char == '-':
result += sign * num
num = 0
sign = -1
elif char == '(':
stack.append(result)
stack.append(sign)
result = 0
sign = 1
num = 0
elif char == ')':
result += sign * num
num = 0
prev_sign = stack.pop()
prev_result = stack.pop()
result = prev_result + prev_sign * result
else:
continue
result += sign * num
return result