Stack implementation calculator

leetcode 224 Use stack to realize calculator .
Extend the , It's not just about adding and subtracting , Multiplication and division are also possible .
In addition, negative numbers can be realized for (-2) + 3 This operation of negative numbers .

Double stack
opv: Digital stack
opc: Symbol stack

Priority words :

1. If the top of the stack is plus or minus , Current is addition and subtraction , You can calculate .
2. If the top of the stack is multiplication and division , Currently, it is not an open parenthesis , You can calculate .
3. If the top of the stack is an open bracket , Currently, if it is a right parenthesis , Just directly Once out of the stack , Otherwise, it doesn't count .

Dealing with negative numbers ：
Negative numbers are not preceded by numbers , There is no right parenthesis , So what satisfies these two characteristics is negative numbers .

The processing of numbers :
Pay attention to the use of numbers in intermediate calculation long Make a strong turn . Otherwise, if it is INT_MAX Explosive int.

Treatment of left and right parentheses :
The right bracket is not put on the stack . The left bracket does not count .

Handling of spaces ：
only Clear the digital counter cur, No calculation .

class Solution {
public:

    int calculate(string s) {
        stack<int>opv;
        stack<char>opc;
        auto cal = [&]() {
            char ch = opc.top(); opc.pop();
            int a = opv.top(); opv.pop();
            int b = opv.top(); opv.pop();
            if (ch == '+') opv.push(b + a);
            else if (ch == '-') opv.push(b - a);
            else if (ch == '*') opv.push(b * a);
            else if (ch == '/') opv.push(b / a);
            // printf("%d %c %d = %d\n", b, ch, a, opv.top());
        };
        auto check = [&](char ch) -> int {
            char top = opc.top();
            if (top == '+' || top == '-') {
                if (ch == '+' || ch == '-' || ch == ')') 
                    return 1;
            }
            else if (top == '*' || top == '/') {
                if (ch != '(')
                    return 1;
            }
            else if (top == '(' && ch == ')') return 0; 
            return -1;
        };
        s += ' ';
        int cur = -1;
        opc.push('#');
        char pre = '#';
        for (char ch : s) {
            if (isdigit(ch)) {
                if (cur == -1) cur = 0;
                cur = (long)cur * 10 + ch - '0';            // 2.  Prevent explosion int
                // printf("%d\n", cur);
            }
            else {
                if (cur != -1) opv.push(cur);               
                cur = -1;
                if (ch == ' ') continue;                    //  Space 
                if (ch == '-' && (!isdigit(pre) && pre != ')')) { //  negative 
                    // printf("pre = %c\n", pre);
                    opv.push(0);
                }
                while (check(ch) > 0) cal();
                if (check(ch) == 0)  //  Right bracket , The left bracket shows a , Not all of them 
                    opc.pop();
                if (ch != ')')             //  The right bracket is not put in 
                    opc.push(ch);
            }
            if (pre != ' ') pre = ch;
        }
        while (opc.top() != '#') cal();
        return opv.top();
    }
};