leetcode 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:

1
2
3

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

Example 2:

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2
3

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

使用栈能够简单的实现

class  {
    public int evalRPN(String[] tokens) {
        Stack<Integer> s = new Stack<>();
        for (String t: tokens){
            if ("+".equals(t)){
                int b = s.pop();
                int a = s.pop();
                int ans = a + b;
                s.push(ans);
            }
            else if ("-".equals(t)){
                int b = s.pop();
                int a = s.pop();
                int ans = a - b;
                s.push(ans);

            }
            else if ("*".equals(t)){
                int b = s.pop();
                int a = s.pop();
                int ans = a * b;
                s.push(ans);
            }
            else if ("/".equals(t)){
                int b = s.pop();
                int a = s.pop();
                int ans = a / b;
                s.push(ans);
            }
            else{
                int num = Integer.parseInt(t);
                s.push(num);
            }
        }
        return s.peek();
    }
}