剑指offer——合并两个排序的链表

 1 /*
 2 public class ListNode {
 3     int val;
 4     ListNode next = null;
 5  
 6     ListNode(int val) {
 7         this.val = val;
 8     }
 9 }*/
10 public class Solution {
11     public ListNode Merge(ListNode list1,ListNode list2) {
12         if(list1==null){
13             return list2;
14         }else if(list2==null){
15             return list1;
16         }
17         ListNode mergeListNode=null;
18         if(list1.val<list2.val){
19             mergeListNode=list1;
20             mergeListNode.next=Merge(list1.next,list2);
21         }else{
22             mergeListNode=list2;
23             mergeListNode.next=Merge(list1,list2.next);
24         }
25         return mergeListNode;
26     }
27 }

 1 public static ListNode merge(ListNode head1, ListNode head2) {
 2         if (head1 == null) {
 3             return head2;
 4         }
 5 
 6         if (head2 == null) {
 7             return head1;
 8         }
 9 
10         ListNode root = new ListNode();
11         // 用于指向合并后的新链的尾结点
12         ListNode pointer = root;
13 
14         // 当两个链表都不为空就进行合并操作
15         while (head1 != null && head2 != null) {
16             if (head1.value < head2.value) {
17                 pointer.next = head1;
18                 head1 = head1.next;
19             } else {
20                 pointer.next = head2;
21                 head2 = head2.next;
22             }
23             pointer = pointer.next;
24         }
25 
26         // 如果第一个链表的元素未处理完将其，接到合并链表的最后一个结点之后
27         if (head1 != null) {
28             pointer.next = head1;
29         }
30 
31         // 如果第二个链表的元素未处理完将其，接到合并链表的最后一个结点之后
32         if (head2 != null) {
33             pointer.next = head2;
34         }
35         return root.next;
36     }