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 .
Example 1:
Input: "1 + 1"
Output: 2
Example 2:
Input: " 2-1 + 2 "
Output: 3
Example 3:
Input: "(1+(4+5+2)-3)+(6+8)"
Output: 23




https://www.youtube.com/watch?v=9c0WHgIsk5g&t=41s
http://www.cnblogs.com/grandyang/p/4570699.html

// use one stack 
// more general : use two stacks, one is for the prev res
// another one is for the sign


Use  one stack 
https://leetcode.com/problems/basic-calculator/discuss/62361/Iterative-Java-solution-with-stack



Use two stacks 



不是我写的， 还有不明白的

class Solution {
    public int calculate(String s) {
        if (s == null || s.length() == 0) return 0;
        Stack<Integer> nums = new Stack<>(); // the stack that stores numbers
        Stack<Character> ops = new Stack<>(); // the stack that stores operators (including parentheses)
        int num = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == ' ') continue;
            if (Character.isDigit(c)) {
                num = c - '0';
                // iteratively calculate each number
                while (i < s.length() - 1 && Character.isDigit(s.charAt(i+1))) {
                    num = num * 10 + (s.charAt(i+1) - '0');
                    i++;
                }
                nums.push(num);
                num = 0; // reset the number to 0 before next calculation
            } else if (c == '(') {
                ops.push(c);
            } else if (c == ')') {
                // do the math when we encounter a ')' until '('
                while (ops.peek() != '(') nums.push(operation(ops.pop(), nums.pop(), nums.pop()));
                ops.pop(); // get rid of '(' in the ops stack
                // ？ 不知道这个precedence 在做什么？ 
            } else if (c == '+' || c == '-' || c == '*' || c == '/') {
                while (!ops.isEmpty() && precedence(c, ops.peek())) nums.push(operation(ops.pop(), nums.pop(),nums.pop()));
                ops.push(c);
            }
        }
        while (!ops.isEmpty()) {
            nums.push(operation(ops.pop(), nums.pop(), nums.pop()));
        }
        return nums.pop();
    }

    private static int operation(char op, int b, int a) {
        switch (op) {
            case '+': return a + b;
            case '-': return a - b;
            case '*': return a * b;
            case '/': return a / b; // assume b is not 0
        }
        return 0;
    }
    // helper function to check precedence of current operator and the uppermost operator in the ops stack 
    // 这个 helper function， 不懂在做什么
    private static boolean precedence(char op1, char op2) {
        if (op2 == '(' || op2 == ')') return false;
        if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-')) return false;
        return true;
    }
}