[LeetCode]-Evaluate Reverse Polish Notation

 1 /*
 2 *解题思路：
 3 *借助辅助栈来完成逆波兰表达式的计算
 4 *1.遍历数组，当前字符串为数字时入栈；
 5 *2.为算符时，弹出栈顶的两个元素，完成计算，计算结果入栈；
 6 *3.遍历完成，当栈非空时，返回栈顶元素，计算完成。
 7 */
 8 public class Solution {
 9     public int evalRPN(String[] tokens) {
10         int factor1, factor2, result=0;
11         Stack<String> stack = new Stack<String>();
12         int len = tokens.length;
13         for( int i=0; i<len; i++ ){
14             if( tokens[i].equals("+") ){
15                 factor1 = Integer.parseInt( stack.pop() );
16                 factor2 = Integer.parseInt( stack.pop() );
17                 result = factor1 + factor2;
18                 stack.push(result+"");
19                 
20             }else if( tokens[i].equals("-") ){
21                 factor1 = Integer.parseInt( stack.pop() );
22                 factor2 = Integer.parseInt( stack.pop() );
23                 result = factor2 - factor1;
24                 stack.push(result+"");
25                 
26             }else if( tokens[i].equals("*") ){
27                 factor1 = Integer.parseInt( stack.pop() );
28                 factor2 = Integer.parseInt( stack.pop() );
29                 result = factor1 * factor2;
30                 stack.push(result+"");
31                 
32             }else if( tokens[i].equals("/") ){
33                 factor1 = Integer.parseInt( stack.pop() );
34                 factor2 = Integer.parseInt( stack.pop() );
35                 result = factor2 / factor1;
36                 stack.push(result+"");
37                 
38             }else{
39                 stack.push( tokens[i] );
40             }
41         }
42         
43         if( !stack.isEmpty() )
44             result = Integer.parseInt( stack.pop() );
45         return result;    
46     }
47 }