为什么链表实现不起作用

huangapple
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英文:

Why is the Linked List implementation not working

问题

在你提供的Java代码中，你试图学习如何实现链表。你创建了一个在链表开头插入元素的方法，但当你通过主函数运行它时，似乎没有看到任何输出返回。

以下是你的代码的翻译部分：

  1. class ListNode{
  2.   int data;
  3.   ListNode next;
  5.   ListNode(int data)
  6.   {
  7.     this.data = data;
  8.     this.next = null;
  9.   }
  10. }
  12. class Operations{
  13.   static ListNode head = null;
  15.   void insertStart(int num, ListNode head)
  16.   {
  17.     ListNode temp = new ListNode(num);
  18.     if (head == null)
  19.     {
  20.       head = temp;
  21.     }
  22.     else{
  23.       temp.next = head;
  24.       head = temp;
  25.     }
  26.   }
  28.   void display(ListNode head){
  29.     ListNode temp = head;
  30.     while (temp != null)
  31.     {
  32.       System.out.println(temp.data + "->");
  33.       temp = temp.next;
  34.     }
  35.   }
  37.   public static void main(String[] args) {
  38.     Operations l1 = new Operations();
  39.     l1.insertStart(5, head);
  40.     l1.insertStart(4, head);
  41.     l1.insertStart(3, head);
  42.     l1.insertStart(2, head);
  43.     l1.insertStart(1, head);
  44.     l1.display(head);
  45.   }
  46. }

请注意，你的插入方法可能存在问题，因为你传递了 head 参数，但似乎没有更新 head。这可能导致你的链表操作不按预期工作。你可以考虑修改插入方法以确保 head 在插入时正确更新。

英文:

So, im trying to learn the implementation of linked list in java. I created a method to insert in the start but when i run it through main, i don't see any output returned.

  1. class ListNode{
  2.   int data;
  3.   ListNode next;
  5.   ListNode(int data)
  6.   {
  7.     this.data = data;
  8.     this.next = null;
  9.   }
  10. }
  12. class Operations{
  13.   static ListNode head = null;
  15.   void insertStart(int num, ListNode head)
  16.   {
  17.     ListNode temp = new ListNode(num);
  18.     if (head == null)
  19.     {
  20.       head = temp;
  21.     }
  22.     else{
  23.       temp.next = head;
  24.       head = temp;
  25.     }
  26.   }
  28.   void display(ListNode head){
  29.     ListNode temp = head;
  30.     while (temp != null)
  31.     {
  32.       System.out.println(temp.data+"->");
  33.       temp = temp.next;
  34.     }
  35.   }
  37.   public static void main(String[] args) {
  38.     Operations l1 = new Operations();
  39.     l1.insertStart(5,head);
  40.     l1.insertStart(4, head);
  41.     l1.insertStart(3, head);
  42.     l1.insertStart(2, head);
  43.     l1.insertStart(1, head);
  44.     l1.display(head);
  45.   }
  46. }

Can someone please help me, i can't seem to find the error. Everything seems perfect.

答案1

得分: 1

问题在于您将head作为参数传递给您的insert方法，因此它是一个局部变量。对局部变量的任何赋值将不会被调用者看到。

但是，退一步来说，将head作为参数传递给链表实例的方法实际上是不必要的。该实例应该知道它的头是什么。这揭示了另一个问题：head不应该是类的静态成员，而应该是一个实例变量：每个Operations的实例都应该有自己的head。

这并不是问题，但insert的实现实际上不需要if...else结构。在任何情况下，都可以将temp.next赋给head，并且在任何情况下，都要设置head为temp。

因此，请按以下方式更正您的代码：

  1. class ListNode {
  2.   int data;
  3.   ListNode next;
  5.   ListNode(int data) {
  6.     this.data = data;
  7.     this.next = null;
  8.   }
  9. }
  11. class Operations {
  12.   ListNode head = null; // 不应该是静态的！每个实例都有自己的head
  14.   void insertStart(int num) // head不应该是参数。使用实例变量。
  15.   {
  16.     ListNode temp = new ListNode(num);
  17.     // 不需要区分空链表或非空链表：
  18.     temp.next = head;
  19.     head = temp;
  20.   }
  22.   void display() { // head不应该是参数。使用实例变量。
  23.     ListNode temp = head;
  24.     while (temp != null) {
  25.       System.out.print(temp.data + "->"); // 在同一行上输出
  26.       temp = temp.next;
  27.     }
  28.     System.out.println("null"); // 终止行
  29.   }
  31.   public static void main(String[] args) {
  32.     Operations l1 = new Operations();
  33.     l1.insertStart(5); // 不要将head作为参数传递。
  34.     l1.insertStart(4);
  35.     l1.insertStart(3);
  36.     l1.insertStart(2);
  37.     l1.insertStart(1);
  38.     l1.display();
  39.   }
  40. }
英文:

The problem is that you pass head as an argument to your insert method, and so it is a local variable. Any assignment to a local variable will not be seen by the caller.

But, taking a step back, it should not be necessary to pass head as argument to a method of your linked list instance. That instance should know "itself" what its head is. And that reveals another problem: head should not be a static member of your class, but an instance variable: every instance of Operations should have its own head.

Not a problem, but the implementation of insert doesn't really need the if...else construct. In either case you can assign temp.next=head, and in either case you want to set head=temp.

So correct your code as follows:

  1. class ListNode{
  2.   int data;
  3.   ListNode next;
  5.   ListNode(int data)
  6.   {
  7.     this.data = data;
  8.     this.next = null;
  9.   }
  10. }
  12. class Operations{
  13.   ListNode head = null; // Should not be static! Every instance has its own head
  15.   void insertStart(int num) // head should not be an argument. Use the instance variable.
  16.   {
  17.     ListNode temp = new ListNode(num);
  18.     // No distinction needed between empty list or non-empty list:
  19.     temp.next = head;
  20.     head = temp;
  21.   }
  23.   void display(){ // head should not be an argument. Use the instance variable.
  24.     ListNode temp = head;
  25.     while (temp != null)
  26.     {
  27.       System.out.print(temp.data+"->"); // Output on same line
  28.       temp = temp.next;
  29.     }
  30.     System.out.println("null"); // Terminate the line
  31.   }
  33.   public static void main(String[] args) {
  34.     Operations l1 = new Operations();
  35.     l1.insertStart(5); // Don't pass the head as argument.
  36.     l1.insertStart(4);
  37.     l1.insertStart(3);
  38.     l1.insertStart(2);
  39.     l1.insertStart(1);
  40.     l1.display();
  41.   }
  42. }

答案2

得分: -1

以下是已翻译的代码部分：

  1. 这是你需要做的事情：
  3. class ListNode {
  4.   int data;
  5.   ListNode next;
  7.   ListNode(int data) {
  8.     this.data = data;
  9.     this.next = null;
  10.   }
  11. }
  13. class Operations {
  14.   static ListNode head = null;
  16.   void insertStart(int num) {
  17.     ListNode temp = new ListNode(num);
  18.     if (head == null) {
  19.       head = temp;
  20.     } else {
  21.       temp.next = head;
  22.       head = temp;
  23.     }
  24.   }
  26.   void display(ListNode head) {
  27.     ListNode temp = head;
  28.     while (temp != null) {
  29.       System.out.println(temp.data + "->");
  30.       temp = temp.next;
  31.     }
  32.   }
  34.   public static void main(String[] args) {
  35.     Operations l1 = new Operations();
  36.     l1.insertStart(5);
  37.     l1.insertStart(4);
  38.     l1.insertStart(3);
  39.     l1.insertStart(2);
  40.     l1.insertStart(1);
  41.     l1.display(head);
  42.   }
  43. }
英文:

Here is what you need to do:

  1.     class ListNode{
  2.   int data;
  3.   ListNode next;
  5.   ListNode(int data)
  6.   {
  7.     this.data = data;
  8.     this.next = null;
  9.   }
  10. }
  12. class Operations{
  13.   static ListNode head = null;
  15.   void insertStart(int num)
  16.   {
  17.     ListNode temp = new ListNode(num);
  18.     if (head == null)
  19.     {
  20.       head = temp;
  21.     }
  22.     else{
  23.       temp.next = head;
  24.       head = temp;
  25.     }
  26.   }
  28.   void display(ListNode head){
  29.     ListNode temp = head;
  30.     while (temp != null)
  31.     {
  32.       System.out.println(temp.data+"->");
  33.       temp = temp.next;
  34.     }
  35.   }
  37.   public static void main(String[] args) {
  38.     Operations l1 = new Operations();
  39.     l1.insertStart(5);
  40.     l1.insertStart(4);
  41.     l1.insertStart(3);
  42.     l1.insertStart(2);
  43.     l1.insertStart(1);
  44.     l1.display(head);
  45.   }
  46. }

huangapple
  • 本文由 发表于 2023年4月4日 14:03:36
  • 转载请务必保留本文链接：https://go.coder-hub.com/75925961.html
  • java
  • linked-list
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