面试题17：合并两个排序的链表

ListNode类（结点类）

 1 package leetcode.utilities;
 2
 3 public class ListNode {
 4     public int    val ;
 5     public ListNode   next ;
 6     public ListNode(int val) {
 7         // TODO Auto-generated constructor stub
 8         this.val = val;
 9         this.next = null ;
10     }
11     //打印链表
12         public static void PrintList(ListNode head){
13             System.out.println("PrinyList  start");
14             ListNode pNode = head ;
15             System.out.print("{ ");
16             while(pNode != null){
17                 System.out.print(pNode.val + " ");
18                 pNode= pNode.next;
19             }
20             System.out.println(" }");
21         }
22 }

递归实现Java代码：

 1 package com.hb.jzoffer;
 2
 3 import leetcode.utilities.ListNode;
 4 //面试题17：合并两个排序的链表
 5 public class MergeSortedLists {
 6     public ListNode merge(ListNode pHead1 , ListNode  pHead2){
 7         //判断两个链表是否为空
 8         if(pHead1 == null ){
 9             return pHead2;
10         }else if(pHead2 == null ){
11             return  pHead1;
12         }
13
14         ListNode  pMergedHead = null;
15
16         //递归调用
17         if(pHead1.val < pHead2.val){
18             pMergedHead = pHead1 ;
19             //下面递归调用自己，返回的值是pMergedHead后面的节点，一直到递归结束，才会返回pMergedHead新的头结点。
20             pMergedHead.next = merge(pHead1.next, pHead2) ;
21         }else{
22             pMergedHead = pHead2 ;
23             pMergedHead.next = merge(pHead1, pHead2.next);
24         }
25
26         return pMergedHead ;
27     }
28
29     public static void main(String[] args) {
30         ListNode head1 = new ListNode(1);
31         ListNode n1 = new ListNode(3);
32         ListNode n2 = new ListNode(5);
33         ListNode n3 = new ListNode(7);
34         ListNode n4 = new ListNode(9);
35
36         head1.next = n1 ;
37         n1.next = n2 ;
38         n2.next = n3 ;
39         n3.next = n4 ;
40
41         ListNode head2 = new ListNode(2);
42         ListNode n5 = new ListNode(4);
43         ListNode n6 = new ListNode(6);
44
45         head2.next = n5 ;
46         n5.next = n6;
47
48         MergeSortedLists  merge = new MergeSortedLists();
49
50         ListNode  mergeHead = merge.merge(head1, head2);
51
52         ListNode.PrintList(mergeHead);
53
54     }
55 }