Leetcode No.150 逆波兰表达式求值（栈）

示例1：
 输入：tokens = ["2","1","+","3","*"]
 输出：9
 解释：该算式转化为常见的中缀算术表达式为：((2 + 1) * 3) = 9
示例2：
 输入：tokens = ["4","13","5","/","+"]
 输出：6
 解释：该算式转化为常见的中缀算术表达式为：(4 + (13 / 5)) = 6
示例3：
 输入：tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
 输出：22
 解释：
 该算式转化为常见的中缀算术表达式为：
   ((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
提示：
 1 <= tokens.length <= 104
 tokens[i] 要么是一个算符（"+"、"-"、"*" 或 "/"），要么是一个在范围 [-200, 200] 内的整数

import java.util.Stack;

public class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> sta=new Stack<Integer>();
        for(int i=0;i<tokens.length;i++){
            System.out.println(tokens[i]);
            if(tokens[i].matches("-{0,1}\\d{1,}")==false){
                Integer a=sta.pop();
                System.out.println("a="+a);
                Integer b=sta.pop();
                System.out.println("b="+b);
                if(tokens[i].equals("+")){
                    sta.push(a+b);
                    System.out.println(sta.peek());
                }
                if(tokens[i].equals("-")){
                    sta.push(b-a);
                }
                if(tokens[i].equals("*")){
                    sta.push(a*b);
                }
                if(tokens[i].equals("/")){
                    sta.push(b/a);
                }
            }else{
                sta.push(Integer.parseInt(tokens[i]));
            }
        }
        return sta.pop();
    }

    public static void main(String[] args) {
        String[] tokens = {"2","1","+","3","*"};
        Solution solution=new Solution();
        System.out.println(solution.evalRPN(tokens));
    }
}