LeetCode 150. Evaluate Reverse Polish Notation

原题链接在这里：https://leetcode.com/problems/evaluate-reverse-polish-notation/

题目：

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

题解：

利用stack, 遇到数字就压栈，遇到运算符就先pop() op2, 再pop() op1, 按op1 运算符op2 计算，得出结果压回栈，最后站内剩下的就是结果.

Time Complexity: O(n). n = tokens.length.

Space: O(n).

AC Java:

 1 class Solution {
 2     public int evalRPN(String[] tokens) {
 3         if(tokens == null || tokens.length == 0){
 4             return 0;
 5         }
 6         
 7         Stack<Integer> stk = new Stack<Integer>();
 8         for(int i = 0; i<tokens.length; i++){
 9             if(tokens[i].equals("+")){
10                 int op2 = stk.pop();
11                 int op1 = stk.pop();
12                 stk.push(op1+op2);
13             }else if(tokens[i].equals("-")){
14                 int op2 = stk.pop();
15                 int op1 = stk.pop();
16                 stk.push(op1-op2);
17             }else if(tokens[i].equals("*")){
18                 int op2 = stk.pop();
19                 int op1 = stk.pop();
20                 stk.push(op1*op2);
21             }else if(tokens[i].equals("/")){
22                 int op2 = stk.pop();
23                 int op1 = stk.pop();
24                 stk.push(op1/op2);
25             }else{
26                 stk.push(Integer.valueOf(tokens[i]));
27             }
28         }
29         
30         return stk.pop();
31     }
32 }

跟上Basic Calculator.