java - 空栈异常

Question

这个 EmptyStackException 继续弹出。显然，我的堆栈中没有任何内容，但用户输入的第一个元素。但是，我不确定代码哪里有缺陷。 (很多地方)但我只需要修复这个错误。

import java.util.*;

public class stacks2 {

public static void main (String []args){
System.out.printf("Enter a math equation in reverse polish notation:\n");

//Create stack of Strings
Stack<String> rpnStack = new Stack<String>();
//Create Scanner 
Scanner input = new Scanner(System.in);
//String in = input.next();

while(input != null) {
    String in = input.next();
        // Tokenize string based on spaces.
        StringTokenizer st = new StringTokenizer(in, " ", true);
            while (st.hasMoreTokens()) {
             rpnStack.push(st.nextToken());
         }
    //Send stack to Calculation Method
    calculate(rpnStack);
     }
}

public static void calculate(Stack<String> stack) {
    // Base case: stack is empty => Error, or finished
    if (!stack.isEmpty())
      // throw new StackUnderflowException("Empty Stack");

    // Base case: stack has 1 element, which is the answer => finished
    if (stack.size() == 1)
        System.out.printf("Finished, Answer: %s\n",stack.peek());

    // Recursive case: stack more elements on it.
    if (stack.size() > 1){
        String temp1 = stack.peek();
        stack.pop();
        String temp2 = stack.peek();
        stack.pop();
        String temp3 = stack.peek();
        stack.pop();


            if (temp3.equals("+")){
            float resultant = Float.parseFloat(temp1) + Float.parseFloat(temp2);
            stack.push(String.valueOf(resultant));
            //System.out.println(resultant);
            calculate(stack);
            }

            if (temp3.equals("-")){
            float resultant = Float.parseFloat(temp1) - Float.parseFloat(temp2);
            stack.push(String.valueOf(resultant)); 
            //System.out.println(resultant);
            calculate(stack);
            }

            else if (temp3.equals("*")){
            float resultant = Float.parseFloat(temp1) * Float.parseFloat(temp2);
            stack.push(String.valueOf(resultant)); 
            //System.out.println(resultant);
            calculate(stack);
            }

            else if (temp3.equals("/")){
            float resultant = Float.parseFloat(temp1) / Float.parseFloat(temp2);
            stack.push(String.valueOf(resultant)); 
            //System.out.println(resultant);
            calculate(stack);
            }

            else{
            System.out.printf("Something severely has gone wrong.");
            }
        }  
    }
}

输入和错误:

:~ Home$ java stacks2
Enter a math equation in reverse polish notation:
4 5 * 6 -
Finished, Answer: 4
Exception in thread "main" java.util.EmptyStackException
at java.util.Stack.peek(Stack.java:85)
at stacks2.calculate(stacks2.java:41)
at stacks2.main(stacks2.java:22)

很明显，这只是第一个元素，这让我觉得我的 while 循环在 17 是原因。有什么见解吗？

Answer 1

String in = input.next(); 读给你一个词，然后你试图标记那个词。也许你的意思是 String in = input.nextLine();

http://docs.oracle.com/javase/1.5.0/docs/api/java/util/Scanner.html#next() http://docs.oracle.com/javase/1.5.0/docs/api/java/util/Scanner.html#nextLine()

---

此外，您的代码中有这两行。

if (!stack.isEmpty())
  // throw new StackUnderflowException("Empty Stack");

这是完全错误的。如果没有花括号，if 会影响下一条语句。这不是评论 - 它是以下如果。

这个:

if (!stack.isEmpty())
// throw new StackUnderflowException("Empty Stack");

// Base case: stack has 1 element, which is the answer => finished
if (stack.size() == 1)
    System.out.printf("Finished, Answer: %s\n",stack.peek());

等同于:

if (!stack.isEmpty())
    if (stack.size() == 1)
        System.out.printf("Finished, Answer: %s\n",stack.peek());

还有这个:

if (!stack.isEmpty() && stack.size() == 1){
    System.out.printf("Finished, Answer: %s\n",stack.peek());
}

道德:始终使用带有if 的花括号并且不要注释掉断言。即使您确实注释掉了断言，也要完整地注释掉它们，而不是其中的一半，尤其是当另一半是没有括号的 if 时。

---

第三，你的逻辑有问题。你这样做:

将所有符号压入堆栈，然后弹出前三个并将它们视为一个运算符和两个数字。这将适用于一些输入如果您改用队列。

4 5 * 6 -

按照您的逻辑，这将弹出 * 6 - 并崩溃。如果您使用队列，它将在这种情况下

起作用

4 5 * 6 - 
20 6 -
14

但不是这种情况:

(1+1)*(1+1)
express as RPN
1 1 + 1 1 + *
2 1 1 + *

接下来，弹出 2 1 1 并崩溃。

相反，你应该做什么:

Read the input. For each symbol:
  if it is a number,
    push it on the stack.
  else,
    pop two numbers from the stack,
    perform the operation and
     push the result.