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

 

题目中并没有考虑 invalid Polish Notation 的处理,还有 Integer Overflow、Devide by 0.

 

Polish Notation(wiki) 的优点:

所有操作符置于操作数的后面,因此也被称为后缀表示法。逆波兰记法不需要括号来标识操作符的优先级.

 

try/catch as conditional:

这里使用 helper 容易扩展更多的运算。但是,这里使用 try/catch 来实现,是非常不恰当的。原因如下:

You really should not use try/catch and if interchangeably.

try/catch is for things that go wrong that are outside of your control and not in the normal program flow. For example, trying to write to a file and the file system is full? That situation should typically be handled with try/catch.

if statements should be normal flow and ordinary error checking. So, for example, user fails to populate a required input field? Use if for that, not try/catch.

 There is more overhead even if the exception is never thrown.

 

Solution with try/catch:

 1 public class Solution {
 2     public int evalRPN(String[] tokens) {
 3         if(tokens == null || tokens.length == 0) {
 4             return 0;
 5         }
 6         
 7         Stack<Integer> stack = new Stack<>();
 8         
 9         for(String str: tokens) {
10             try {
11                 int num = Integer.parseInt(str);
12                 stack.push(num);
13             } catch (Exception e){
14                 int num2 = stack.pop();
15                 int num1 = stack.pop();
16                 int num = helper(num1, num2, str);
17                 stack.push(num);
18             }
19         }
20         
21         return stack.pop();
22     }
23     
24     private int helper(int num1, int num2, String str) {
25         int res = 0;
26         
27         switch(str) {
28             case "*":
29                 res = num1 * num2;
30                 break;
31             case "/":
32                 res = (num2 == 0) ? 0: num1 / num2;
33                 break;
34             case "+":
35                 res = num1 + num2;
36                 break;
37             case "-":
38                 res = num1 - num2;
39         }
40         return res;
41     }
42 }

 

Solution without try/catch:

 

public class Solution {
    public int evalRPN(String[] tokens) {
        if(tokens == null || tokens.length == 0) {
            return 0;
        }
        
        Stack<Integer> stack = new Stack<>();
        
        for(String str: tokens) {
            switch(str) {
                case "*":
                    stack.push(stack.pop() * stack.pop());
                    break;
                case "+":
                    stack.push(stack.pop() + stack.pop());
                    break;
                case "-":
                    stack.push(-stack.pop() + stack.pop());
                    break;
                case "/":
                    int num1 = stack.pop();
                    int num2 = stack.pop();
                    int num = (num1 == 0)?0:num2/num1;
                    stack.push(num);
                    break;
                default:
                    stack.push(Integer.parseInt(str));
            }
        }
        
        return stack.pop();
    }
}

 

posted @ 2016-01-04 19:41  一切不期而至  阅读(196)  评论(0编辑  收藏  举报