逆波兰表示法

题目描述：

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
思路：利用栈

解法一：
import java.util.*;
public class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> s=new Stack<Integer>();
        for(int i=0;i<tokens.length;i++){
            switch(tokens[i]){
                case "+":
                    s.push(s.pop()+s.pop());
                    break;
                case "-":
                    int num1=s.pop();
                    int num2=s.pop();
                    s.push(num2-num1);
                    break;
                case "*":
                    s.push(s.pop()*s.pop());
                    break;
                case "/":
                    int num3=s.pop();
                    int num4=s.pop();
                    s.push(num4/num3);
                    break;
                default:
                    s.push(Integer.parseInt(tokens[i]));
            }
        }
        return s.pop();
    }
}

解法二：
import java.util.*;
public class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> stack=new Stack<Integer>();
        for(int i=0;i<tokens.length;i++){
            try{
                int num=Integer.parseInt(tokens[i]);
                stack.add(num);
            }catch(Exception e){
                int b=stack.pop();
                int a=stack.pop();
                stack.add(get(a,b,tokens[i]));
            }
        }
        return stack.pop();
    }
    private int get(int a,int b,String operator){
        switch(operator){
            case "+":
                return a+b;
            case "-":
                return a-b;
            case "*":
                return a*b;
            case "/":
                return a/b;
            default:
                return 0;
        }
    }
}