772. Basic Calculator III

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

The expression string contains only non-negative integers, +, -, *, / operators , open ( and closing parentheses ) and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-2147483648, 2147483647].

Some examples:

"1 + 1" = 2
" 6-4 / 2 " = 4
"2*(5+5*2)/3+(6/2+8)" = 21
"(2+6* 3+5- (3*14/7+2)*5)+3"=-12

想法跟227. Basic Calculator II很類似，最大的不同是利用recursion處理括號內的運算。

int calculate(string s) { // time: O(n); space: O(n)
    int n = s.length(), curRes = 0, res = 0;
    long num = 0;
    char op = '+';
    for (int i = 0; i < n; ++i) {
        char c = s[i];
        if (c >= '0' && c <= '9') num = num * 10 + (c - '0');
        else if (c == '(') {
            int j = i, cnt = 0;
            for (; i < n; ++i) {
                if (s[i] == '(') ++cnt;
                else if (s[i] == ')') --cnt;
                if (cnt == 0) break;
            }
            num = calculate(s.substr(j + 1, i - j - 1));
        }
        if (c == '+' || c == '-' || c == '*' || c == '/' || i == n - 1) {
            switch (op) {
                case '+': curRes += num; break;
                case '-': curRes -= num; break;
                case '*': curRes *= num; break;
                case '/': curRes /= num; break;
            }
            if (c == '+' || c == '-' || i == n - 1) {
                res += curRes;
                curRes = 0;
            }
            op = c;
            num = 0;
        }
    }
    return res;
}