连接两个链表

Question

编写一个concatenate((方法，该方法接受两个链表list1和list2，并将list2附加到list1的末尾。

例如，如果列表最初是：列表1:

head-->3-->6-->7-->2-->null

列表2:

head-->9-->5-->3-->null

list1在串联后变为：

head-->3-->6-->7-->2-->9-->5-->3-->null
public static void concatenate(LinkedList list1, LinkedList list2) {
//code

}

Answer 1

想法：您需要一个指针来指向list1的头部。然后将指针移动到list1的最后一个元素，并将其分配到list2头部旁边的点。

public class Node  
{ 
int data; 
Node next; 
Node(int d) {data = d; 
next = null;} 
} 

public static Node concatenate (Node head1, Node head2) //head1 points to head of list1, head2 points to head of list2
{
Node temp=null;
if (head1==NULL) //if the first linked list is empty
return (head2);
if (head2==NULL) //if second linked list is empty
return (head1);
temp=head1;       //place temporary pointer on the first node of the first linked list
while (temp.next!=NULL) //move p to the last node
temp=temp.next;
temp.next=head2;                           //address of the first node of the second linked list stored in the last node of the first linked list
return (head1);
}