java - 使用堆栈测试字符串是否为回文

Question

我正在尝试获取用户输入的单词，将其添加到堆栈中，然后检查该单词是否是回文。我试图将所有内容从堆栈中弹出并放到一个新字符串上，然后比较字符串。我相信我当前的问题是我的 pop() 函数实际上并没有返回值。它实际上只是切断了尾节点，然后重新打印没有尾节点的字符串。我想我的问题是，如何编写 pop() 函数以便它返回“弹出”值？

这是我的主要方法

 Scanner input = new Scanner(System.in);
     Stack_Scott_Robinson<Character> myList = new Stack_Scott_Robinson<Character>(); //create a list object
     System.out.println("Enter a string: ");
     String s = input.next();
     for (int i = 0; i < s.length(); i++){
        myList.push(s.charAt(i));
     }
     String reverseInput = "";
     while (!myList.isEmpty()){
        reverseInput = reverseInput + myList.Pop();
     }
     if (s.equals(reverseInput)){
        System.out.println("This is a palindrome");
     }else{
        System.out.println("This is not a palidrome");
     System.out.print("Would you like to re-run code with different input string(y/n)?");
     another = input.next();
     }

这是我的 pop() 方法

public void Pop()
{
  Node countList = end; //set "count" node at the front of the list
  int size = 0; //initialize the size variable
  //moves the count node down the list and increments the size variable
  while (countList != null)
  {
     countList = countList.next;
     size++;
  }
  if (size == 0){ //empty list
     System.out.println("error: empty list");
  }else if (size == 1) //one node list
  {
     top = null;
     end = null;
     System.out.println("list is now empty");
  }
  else{
     Node current = end;//set a current node at the front of the list
     for (int i = 0; i<size-2; i++)//-2 because the list starts at 0 and we want the second to last position
     {
        current = current.next;//moves the current node down the list until it is 
     } 
     Node temporary = current.next;  //the second to last position
     Node temp = top;//create a new node space
     top = current;//set the new node equal to the second to last position
     top.next = null;//sets the node as the tail
  }
}   

Answer 1

这是 java.util.Vector 的工作原理(java.util.Stack 扩展了 Vector)

public synchronized E pop() {
    E       obj;
    int     len = size();

    obj = peek();
    removeElementAt(len - 1);

    return obj;
}

和

public synchronized void removeElementAt(int index) {
    if (index >= elementCount) {
        throw new ArrayIndexOutOfBoundsException(index + " >= " +
                                                 elementCount);
    }
    else if (index < 0) {
        throw new ArrayIndexOutOfBoundsException(index);
    }
    int j = elementCount - index - 1;
    if (j > 0) {
        System.arraycopy(elementData, index + 1, elementData, index, j);
    }
    modCount++;
    elementCount--;
    elementData[elementCount] = null; /* to let gc do its work */
}

在 pop 方法中，您可以看到我们知道要删除哪个对象，因为他们使用 peek() 获取它；然后他们只是删除 vector 末尾的元素。希望这有帮助。我不确定你在幕后如何代表你的堆栈，所以如果没有这个，我真的无法给你更好的答案。