Basic Calculator LeetCode Solution

Problem – Basic Calculator LeetCode Solution

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.

Basic Calculator LeetCode Solution in Java

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;
}

Basic Calculator LeetCode Solution in C++

int calculate(string s) {
    int total = 0;
    vector<int> signs(2, 1);
    for (int i=0; i<s.size(); i++) {
        char c = s[i];
        if (c >= '0') {
            int number = 0;
            while (i < s.size()  &&  s[i] >= '0')
                number = 10 * number + s[i++] - '0';
            total += signs.back() * number;
            signs.pop_back();
            i--;
        }
        else if (c == ')')
            signs.pop_back();
        else if (c != ' ')
            signs.push_back(signs.back() * (c == '-' ? -1 : 1));
    }
    return total;
}

Basic Calculator LeetCode Solution in Python

def calculate(self, s):
    total = 0
    i, signs = 0, [1, 1]
    while i < len(s):
        c = s[i]
        if c.isdigit():
            start = i
            while i < len(s) and s[i].isdigit():
                i += 1
            total += signs.pop() * int(s[start:i])
            continue
        if c in '+-(':
            signs += signs[-1] * (1, -1)[c == '-'],
        elif c == ')':
            signs.pop()
        i += 1
    return total
Basic Calculator LeetCode Solution Review:

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This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

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