224. Basic Calculator¶

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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 * 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.

Solution¶

class Solution {
    public static int calculate(String s) {
        int len = s.length(), sign = 1, result = 0;
        Stack<Integer> stack = new Stack<Integer>();
        for (int i = 0; i < len; i++) {
            if (Character.isDigit(s.charAt(i))) {
                int sum = s.charAt(i) - '0';
                while (i + 1 < len && Character.isDigit(s.charAt(i + 1))) {
                    sum = sum * 10 + s.charAt(i + 1) - '0';
                    i++;
                }
                result += sum * sign;
            } 
            else if (s.charAt(i) == '+') sign = 1;
            else if (s.charAt(i) == '-') sign = -1;
            else if (s.charAt(i) == '(') {
                stack.push(result);
                stack.push(sign);
                result = 0;
                sign = 1;
            } 
            else if (s.charAt(i) == ')') result = result * stack.pop() + stack.pop();
        }
        return result;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here