Last updated on July 20th, 2024 at 04:10 am

Here, We see Basic Calculator LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Math, Stack

Companies

Google

Level of Question

Hard

Basic Calculator LeetCode Solution

Problem Statement

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

1. Basic Calculator Solution C++

class Solution {
public:
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;
}
};

2. Basic Calculator Solution Java

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

3. Basic Calculator Solution JavaScript

var calculate = function(s) {
    let sign = 1, sum = 0;
    const stack = []; 
    for (let i = 0; i < s.length; i += 1) {
        if (s[i] >= '0' && s[i] <= '9') {
            let num = 0
            while (s[i] >= '0' && s[i] <= '9') {
                num = (num * 10) + (s[i] - '0');
                i += 1;
            }
            sum += (num * sign);
            i -= 1;
        } else if (s[i] === '+') {
            sign = 1;
        } else if (s[i] === '-') {
            sign = -1;
        } else if (s[i] === '(') {
            stack.push(sum);
            stack.push(sign);
            sum = 0
            sign = 1;
        } else if (s[i] === ')') {
            sum = stack.pop() * sum;
            sum += stack.pop();
        }
    }
    return sum;
};

4. Basic Calculator Solution Python

class Solution(object):
    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