Basic Calculator

题目
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 .

答案

class Solution {
    Stack numbers = new Stack<>();
    Stack ops = new Stack<>();

    public int calculate(String s) {
        String num = "";

        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if(c == ' ') continue;
            if(Character.isDigit(c)) {
                num = num + c;
                // If no next char, or next char is not digit, then this is the number
                if(i == s.length() - 1 || !Character.isDigit(s.charAt(i + 1))) {
                    numbers.push(Long.parseLong(num));
                    num = "";
                }
            }
            else if(c == '(') {
                ops.push(c);
            }
            else if(c == ')') {
                // Keep calculating until we see (
                while(ops.peek() != '(') {
                    char op = ops.peek();
                    long second = numbers.pop();
                    long first = numbers.pop();
                    operation(ops.pop(), first, second);
                }
                ops.pop();
            }
            else {
                if(!ops.empty() && ops.peek() != '(') {
                    long second = numbers.pop();
                    long first = numbers.pop();
                    operation(ops.pop(), first, second);
                }

                ops.push(c);
            }
        }

        while(numbers.size() > 1) {
            long second = numbers.pop();
            long first = numbers.pop();
            operation(ops.pop(), first, second);
        }
        return numbers.pop().intValue();
    }

    public void operation(char op, long first, long second) {
        if(op == '+') numbers.push(first + second);
        if(op == '-') numbers.push(first - second);
    }
}