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: 9Explanation: ((2 + 1) * 3) = 9

Example 2:

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

Example 3:

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

难度：medium

题目：

求算术表达式在反波兰表示法中的值。有效的操作符是+、-、*、/。每个操作数可以是一个整数或另一个表达式。

思路：栈

Runtime: 6 ms, faster than 91.00% of Java online submissions for Evaluate Reverse Polish Notation.

Memory Usage: 36.6 MB, less than 100.00% of Java online submissions for Evaluate Reverse Polish Notation.

class Solution {    public int evalRPN(String[] tokens) {        Stack
    
      stack = new Stack<>();        int tNum = 0;        for (int i = 0; i < tokens.length; i++) {            switch(tokens[i]) {                case "+" :                    stack.push(stack.pop() + stack.pop());                    break;                case "-" :                    tNum = stack.pop();                    stack.push(stack.pop() - tNum);                    break;                case "*" :                    stack.push(stack.pop() * stack.pop());                    break;                case "/" :                    tNum = stack.pop();                    stack.push(stack.pop() / tNum);                    break;                default :                    stack.push(Integer.parseInt(tokens[i]));            }        }                return stack.pop();    }}