题解 | #后缀表达式求值#

注意点：
1.减法和除法都是减（除）数先入栈，被减（除）数后入栈，因此是op2-(/)op1

代码块
import java.util.*;


public class Solution {
    /**
     * 
     * @param tokens string字符串一维数组 
     * @return int整型
     */
    public int evalRPN (String[] tokens) {
        // write code here
        Stack <Integer> str = new Stack<Integer>();
        int sum;
        int op1 = 0;
        int op2 = 0;
        for(int i = 0; i < tokens.length; i++){
            switch(tokens[i]){
                case "+":
                    op1 = str.pop();
                    op2 = str.pop();
                    str.push(op1+op2);
                    break;
                case "-":
                    op1 = str.pop();
                    op2 = str.pop();
                    str.push(op2-op1);
                    break;
                case "*":
                    op1 = str.pop();
                    op2 = str.pop();
                    str.push(op1*op2);
                    break;
                case "/":
                    op1 = str.pop();
                    op2 = str.pop();
                    str.push(op2/op1);
                    break;
                default:
                    str.push(Integer.parseInt(tokens[i]));
            }
        }
        return str.peek();
    }
}