Combining two sorted linked lists of java programming problems

In this paper, we share the linked list of java merging two sorts for your reference. The specific contents are as follows

/**
 * 
 *  Sword finger offer Programming problem ( JAVA Realization ) -No. 1 16 Question: Merging two sorted linked lists 
 * 
 *  Input two monotonically increasing linked lists, output the linked list after synthesis of two linked lists,   Of course, we need the synthesized linked list to satisfy the monotone unabated rule. 
 * 
 */
public class Test16 {
 public static ListNode Merge(ListNode list1, ListNode list2) {
 if (list1 == null) { //  First, judge whether a linked list is empty 
 return list2;
 } else if (list2 == null) {
 return list1;
 }
 ListNode end1 = list1;
 ListNode end2 = list2;
 ListNode tmp; //end1 And end2 Represent two linked lists respectively, tmp Used for intermediate synthesis linked list 
 
 if (end1.val > end2.val) {// Think of a linked list with a small first node as end1
 tmp = end1;
 end1 = end2;
 end2 = tmp;
 } else {

 }
 ListNode newNode = end1;// The first node of the linked list for final return 

 while (end1.next != null && end2.next != null) { // To link the list 2 Insert elements in the linked list 1 Appropriate position in 
 if (end1.val <= end2.val && end1.next.val >= end2.val) {
 tmp = end2.next;
 end2.next = end1.next;
 end1.next = end2;
 end1 = end2;
 end2 = tmp;
 } else {
 end1 = end1.next;
 }
 }
 
 if (end1.next == null) {// If linked list 1 When you reach the tail node, you will directly connect the remaining linked list 2 The first node in the 
 end1.next = end2;
 return newNode;
 } else {
 if (end1.next != null && end2.next == null) {// If linked list 2 When you reach the tail node, you will link the list 2 The last remaining in the 1 Insert nodes into linked list 1
 while (end2 != null) {
  if (end1.val <= end2.val && end1.next.val >= end2.val) {
  end2.next = end1.next;
  end1.next = end2;
  break;
  } else {
  end1 = end1.next;
  if (end1.next == null) {// Linked list 2 The last node is the largest 
  end1.next = end2;
  break;
  }
  }
 }
 }
 return newNode;
 }
 }

 public static void main(String[] args) {
 ListNode list1 = new ListNode(1);
 list1.next = new ListNode(3);
 list1.next.next = new ListNode(5);
 ListNode list2 = new ListNode(2);
 list2.next = new ListNode(4);
 list2.next.next = new ListNode(6);
 System.out.println(Merge(list2, list1));
 }

 //  Linked list 
 public static class ListNode {
 int val;
 ListNode next = null;

 ListNode(int val) {
 this.val = val;
 }
 }
}