150. Evaluate Reverse Polish Notation

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Note:

• Division between two integers should truncate toward zero.
• The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.

Example 1:

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation: 
  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

class Solution {
    public int evalRPN(String[] tokens) {
        Stack<String> s = new Stack<>();
        for (String token : tokens) {
            if (!isOperator(token)) {
                s.push(token);
            } else {
                int y = Integer.parseInt(s.pop());
                int x = Integer.parseInt(s.pop());
                switch (token.charAt(0)) {
                    case '+': x += y; break;
                    case '-': x -= y; break;
                    case '*': x *= y; break;
                    default: x /= y;
                }
                s.push(String.valueOf(x));
            }
        }
        return Integer.parseInt(s.peek());
    }
    private static boolean isOperator(final String op) {
        return op.length() == 1 && OPS.indexOf(op) != -1;
    }
    private static String OPS = new String("+-*/");
}

想起了宝拉老哥，用stack解决，遇到数字就push，遇到标点就pop两个数字，后pop的是被除数，运算完成后再push回stack。

最后peek顶端的。

检查是不是标点符号有点东西，用index来判断。