1+ /*
2+ Author: Andy, nkuwjg@gmail.com
3+ Date: Jan 31, 2015
4+ Problem: Evaluate Reverse Polish Notation
5+ Difficulty: Easy
6+ Source: https://oj.leetcode.com/problems/evaluate-reverse-polish-notation/
7+ Notes:
8+ Evaluate the value of an arithmetic expression in Reverse Polish Notation.
9+ Valid operators are +, -, *, /. Each operand may be an integer or another expression.
10+ Some examples:
11+ ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
12+ ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
13+ Solution: stack.
14+ */
15+
16+ public class Solution {
17+ public int evalRPN (String []tokens ) {
18+ Stack <Integer >stk =new Stack <Integer >();
19+ for (int i =0 ;i <tokens .length ; ++i ) {
20+ if ((tokens [i ].compareTo ("+" ) !=0 ) && (tokens [i ].compareTo ("-" ) !=0 )
21+ && (tokens [i ].compareTo ("*" ) !=0 ) && (tokens [i ].compareTo ("/" ) !=0 )) {
22+ stk .push (Integer .parseInt (tokens [i ]));
23+ }else {
24+ int op2 =stk .pop ();
25+ int op1 =stk .pop ();
26+ int res =0 ;
27+ if (tokens [i ].compareTo ("+" ) ==0 )
28+ res =op1 +op2 ;
29+ else if (tokens [i ].compareTo ("-" ) ==0 )
30+ res =op1 -op2 ;
31+ else if (tokens [i ].compareTo ("*" ) ==0 )
32+ res =op1 *op2 ;
33+ else if (tokens [i ].compareTo ("/" ) ==0 )
34+ res =op1 /op2 ;
35+ stk .push (res );
36+ }
37+ }
38+ return stk .pop ();
39+ }
40+ }
