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().

Test Cases

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 * 10<sup>5</sup>
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.

Solution

        
class Solution {
public:
    int calculate(string s) {
        stack<int> st;
        int result = 0, num = 0, sign = 1;
        st.push(sign);
        for(int i=0; i<s.length(); i++) {
            char c = s[i];
            if (c >= '0' && c <= '9') {
                num = num*10 - (int)'0' + (int) c;
            } else if (c == '+' || c == '-') {
                result += sign * num;
                sign = st.top() * (c == '+' ? 1: -1);
                num = 0;
            } else if (c == '(') {
                st.push(sign);
            } else if (c == ')'){
                st.pop();
            }
        }
        result += sign * num;
        return result;
    }
};

    

        
class Solution {
    public int calculate(String s) {
        int result = 0, sign = 1, num = 0;
        Stack<Integer> stack = new Stack<Integer>();
        stack.push(sign);
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if(c >= '0' && c <= '9') {
                num = num * 10 + (c - '0');
            } else if(c == '+' || c == '-') {
                result += sign * num;
                sign = stack.peek() * (c == '+' ? 1: -1);
                num = 0;
            } else if(c == '(') {
                stack.push(sign);
            } else if(c == ')') {
                stack.pop();
            }
        }
        result += sign * num;
        return result;
    }
}

    

        
class Solution:
    def calculate(self, s: str) -> int:
        res, num, sign = 0, 0, 1
        stack = [sign]
        for c in s:
            if '0' <= c <= '9':
                num = num * 10 + (ord(c) - ord('0'))
            elif c in ['+', '-']:
                res += sign * num
                sign = stack[-1] * (-1 if c == '-' else 1)
                num = 0
            elif c == '(':
                stack.append(sign)
            elif c == ')':
                stack.pop()
        res += sign * num
        return res

    

Basic Calculator

Test Cases

Solution

Time Complexity: O(n)

Space Complexity: O(n)