简单后缀表达式求值

• 后缀表达式求值问题

Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are+,-,*,/. Each operand may be an integer or another expression.

• Some examples:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9

  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

• 代码

import java.util.*;

public class Solution {

    public static HashSet opSet = new HashSet();

    static{

        opSet.add("+");

        opSet.add("-");

        opSet.add("*");

        opSet.add("/");

    }

    public int evalRPN(String[] tokens) {

        Stack numStack = new Stack<>();

        for(int i = 0 ;i < tokens.length; i ++){

            if(opSet.contains(tokens[i])){

                //是操作符

                Integer num1;

                Integer num2;

                switch(tokens[i]){

                    case "+" :

                        num1 = numStack.pop();

                        num2 = numStack.pop();

                        numStack.push(num1 + num2);

                        break;

                    case "-" :

                        num1 = numStack.pop();

                        num2 = numStack.pop();

                        numStack.push(num2 - num1);

                        break;

                    case "*" :

                        num1 = numStack.pop();

                        num2 = numStack.pop();

                        numStack.push(num1 * num2);

                        break;

                    case "/" :

                        num1 = numStack.pop();

                        num2 = numStack.pop();

                        numStack.push(num2 / num1);

                        break;

                    default:

                        break;

                }

            }else{

                //是操作数

                numStack.push(Integer.parseInt(tokens[i]));

            }

        }

        return numStack.pop();

    }

}