Question

我正在尝试编写代码以合并两个已排序的列表和另一个列表中的显示。

我创建了一个函数“Merge”，它将每个List的头部作为它的两个Arguments并返回新List的头部。我创建了“Display”函数，它使用“list of the list”来显示其中的Contents 。问题是当我试图显示新List的内容时，它显示Nothing。这是我的主要功能。

public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Linkedlist o=new Linkedlist();
        Linkedlist o1=new Linkedlist();
        Linkedlist o2=new Linkedlist();

        o.insertAtlast(5);
        o.insertAtlast(10);
        o.insertAtlast(15);

        o1.insertAtlast(2);
        o1.insertAtlast(3);
        o1.insertAtlast(20);

        o2.head=o2.mergeList(o.head, o1.head);
        o2.Display(o2.head);

    }

}

and this is my Linkedlist Class

public class Linkedlist {

    static class Node{
        int data;
        Node link;
        public Node(int data) {
            this.data=data;
            this.link=null;
        }
    }

    Node head=null;

    public void insertAtlast(int data) {
        Node node=new Node(data);

        if(head==null)
            head=node;
        else
        {
            Node ptr=head;

            while(ptr.link!=null)
                ptr=ptr.link;
            ptr.link=node;
        }


    }


    public void Display(Node node) {
        while(node!=null) {
            System.out.println(node.data);
            node=node.link;
        }
    }




    public Node mergeList(Node head1,Node head2) {
        Node head3=null;

        if(head1==null)
            head=head2;
        else if(head2==null)
                head=head1;
        else {
            while(head1!=null && head2!=null) {

                if(head==null) {
                    if(head2.data<head1.data) {
                        Node node=new Node(head2.data);
                        head=node;
                        head2=head2.link;
                    }
                    else {
                        Node node=new Node(head1.data);
                        head=node;
                        head1=head1.link;
                    }
                }

                else
                {
                  head3=head;
                  while(head3.link!=null)
                      head=head3.link;
                  if(head2.data<head1.data) {
                      Node node=new Node(head2.data);
                      head3.link=node;
                      head2=head2.link;
                  }
                  else {
                      Node node=new Node(head1.data);
                      head3.link=node;
                      head1=head1.link;
                  }
                }

            }

        }

        return head;

    }

}

Answer 1

递归合并：

public Node mergeList(Node head1, Node head2) {
    if (head1 == null) return head2;
    if (head2 == null) return head1;

    if (head1.data < head2.data) {
        head1.link = mergeList(head1.link, head2);
        return head1;
    } else {
        head2.link = mergeList(head2.link, head1);
        return head2;
    }
}

迭代合并：

public Node mergeList(Node head1, Node head2) {

    Node head3 = null, cursor = null, cursorH1 = head1, cursorH2 = head2;

    while (cursorH1 != null && cursorH2 != null) {
        Node temp;

        if (cursorH1.data < cursorH2.data) {
            temp = cursorH1;
            cursorH1 = cursorH1.link;
        } else {
            temp = cursorH2;
            cursorH2 = cursorH2.link;
        }

        if (head3 == null) {
            cursor = temp;
            head3 = cursor;
        } else {
            cursor.link = temp;
            cursor = cursor.link;
        }
    }

    if (cursorH1 != null) {
        if (head3 == null) {
            head3 = cursorH1;
        } else {
            cursor.link = cursorH1;
        }
    }

    if (cursorH2 != null) {
        if (head3 == null) {
            head3 = cursorH2;
        } else {
            cursor.link = cursorH2;
        }
    }

    return head3;
}

合并两个已排序的Lnkedlist并在新的链接列表中打印

1 个答案: