Evaluate a mathematical expression given as a string.
Given a string representing a mathematical expression, evaluate it and return its value. The expression can contain basic arithmetic operators, and integer division should truncate toward zero.
Input: s = "3+5/2"
Output: 5
Explanation: Following the order of operations (PEMDAS), 5 / 2 = 2 (integer division), and 3 + 2 = 5.

Question

Evaluate a mathematical expression given as a string.

Given a string representing a mathematical expression, evaluate it and return its value. The expression can contain basic arithmetic operators, and integer division should truncate toward zero.

Input: s = "3+5/2"

Output: 5

Explanation: Following the order of operations (PEMDAS), 5 / 2 = 2 (integer division), and 3 + 2 = 5.

Anonymous · Accepted Answer

// O(N) time. Uses variables to track the last number instead of a full stack to save space.
public int calculate(String s) {
    int len = s.length(), num = 0, res = 0, lastNum = 0;
    char sign = '+';
    for (int i = 0; i < len; i++) {
        char c = s.charAt(i);
        if (Character.isDigit(c)) num = num * 10 + (c - '0');
        if (!Character.isDigit(c) && c != ' ' || i == len - 1) {
            if (sign == '+' || sign == '-') {
                res += lastNum;
                lastNum = (sign == '+') ? num : -num;
            } else if (sign == '*') lastNum *= num;
            else if (sign == '/') lastNum /= num;
            sign = c; num = 0;
        }
    }
    return res + lastNum;
}

Anonymous · Answer

we will have one 1 stack for parenthesis and expressions and one stack for
numbers/digits. We start reading the string haracters one by one and push it to
the relevant stacks.

as soon as finish reading the string. we start rocessing by popping from the
stack-1 which has the expressions and parenthesis. when we hit an expression and
pop it we pop 2 digits from the other stack continuously.

Evaluate a mathematical expression given as a string.

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