java - 如何合并2个链表

我试图在我的编程课上提出一种方法来合并两个链表，以完成一项作业。我在这里真的很困惑，该方法必须具有以下方法签名：

public UnorderedLinkedListInt merge2(UnorderedLinkedListInt list)，因此在我的测试器方法中，它看起来像这样的list3 = list1.merge2(list2)。当方法仅包含一个列表而不同时包含两个列表时，我对如何进行此设置感到困惑。到目前为止，这是我的代码

    public class UnorderedLinkedListInt extends LinkedListIntClass {
    //Default constructor
    public UnorderedLinkedListInt() {
        super();
    }

    public boolean search(int searchItem)  {
        LinkedListNode current; //variable to traverse the list
        current = first;
        while (current != null)
            if (current.info == searchItem)
                return true;
            else
               current = current.link;
        return false;
    }

    public void insertFirst(int newItem) {
        LinkedListNode newNode;  //variable to create the new node
        //create and insert newNode before first
        newNode = new LinkedListNode(newItem, first);
        first = newNode;
        if (last == null)
            last = newNode;
        count++;
    }

    public void insertLast(int newItem)  {
        LinkedListNode newNode; //variable to create the new node
        //create newNode
        newNode = new LinkedListNode(newItem, null);
        if (first == null) {
            first = newNode;
            last = newNode;
        }
        else {
            last.link = newNode;
            last = newNode;

        }
        count++;
    }

    public void deleteNode(int deleteItem) {
        LinkedListNode current; //variable to traverse the list
        LinkedListNode trailCurrent; //variable just before current
        boolean found;
        //Case 1; the list is empty
        if ( first == null)
            System.err.println("Cannot delete from an empty list.");
        else {
            //Case 2: the node to be deleted is first
            if (first.info == deleteItem) {
                first = first.link;
                if (first == null)  //the list had only one node
                    last = null;
                count--;
            }
            else {  //search the list for the given info
                found = false;
                trailCurrent = first; //trailCurrent points to first node
                current = first.link; //current points to second node
                while (current != null && !found) {
                    if (current.info == deleteItem)
                        found = true;
                    else {
                        trailCurrent = current;
                        current = current.link;
                    }
                }
                //Case 3; if found, delete the node
                if (found) {
                    count--;
                    trailCurrent.link = current.link;
                    if (last == current)  //node to be deleted was the last node
                       last = trailCurrent;
                }
                else
                   System.out.println("Item to be deleted is not in the list.");
            }
        }
    }
    public void merge(UnorderedLinkedListInt list2){
      UnorderedLinkedListInt list1 = this;

        while (list2.first != null) {//while more data to print
            list1.insertLast(list2.first.info);
            list2.first = list2.first.link;
        }
    }
    public UnorderedLinkedListInt merge2(UnorderedLinkedListInt list2){
      UnorderedLinkedListInt list3 = new UnorderedLinkedListInt();
      UnorderedLinkedListInt list1 = this;
      while (list1.first != null) {//while more data to print
            list3.insertLast(list1.first.info);
            list1.first = list1.first.link;
        }
      while (list2.first != null) {//while more data to print
            list3.insertLast(list2.first.info);
            list2.first = list2.first.link;
        }
        return list3;
    }
}

我仍然在准确地了解链表的工作方式方面遇到一些麻烦，关于如何设计此方法的任何建议将不胜感激。

最佳答案

在像list1.merge2(list2)这样的方法调用中，该方法将list1接收为隐式“当前对象”，您可以使用this引用对其进行访问。

如果愿意，可以使用其他名称：

public UnorderedLinkedListInt merge2(UnorderedLinkedListInt list2){
    UnorderedLinkedListInt list1 = this;
    // now merge list1 and list2
}