您的位置:首页 > 编程语言 > Lua

150. Evaluate Reverse Polish Notation

2017-06-13 10:35 288 查看
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

链接: http://leetcode.com/problems/evaluate-reverse-polish-notation/

6/13/2017

15ms, 42%

用stack来保存之前的值,注意最后一步返回stack的剩余元素

1 public class Solution {
2     public int evalRPN(String[] tokens) {
3         if (tokens == null) {
4             return 0;
5         }
6         Stack<Integer> stack = new Stack<Integer>();
7         int result = 0;
8         for (int i = 0; i < tokens.length; i++) {
9             if (tokens[i].equals("+") || tokens[i].equals("-") || tokens[i].equals("*") || tokens[i].equals("/")) {
10                 int operand2 = stack.pop();
11                 int operand1 = stack.pop();
12                 if (tokens[i].equals("+")) {
13                     result = operand1 + operand2;
14                 } else if (tokens[i].equals("-")) {
15                     result = operand1 - operand2;
16                 } else if (tokens[i].equals("*")) {
17                     result = operand1 * operand2;
18                 } else {
19                     result = operand1 / operand2;
20                 }
21                 stack.push(result);
22             } else {
23                 stack.push(Integer.parseInt((tokens[i])));
24             }
25         }
26         return stack.pop();
27     }
28 }


更多讨论

https://discuss.leetcode.com/category/158/evaluate-reverse-polish-notation
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: