题目

这道题是链表的简单应用，将两个有序链表合成一个有序链表。

思路是：表一，表二各取两个对象，分别指向current和next，进行交叉比较排序。

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
      //判断三种null情况
       if(l1==null && l2==null) return l1;
       if(l1==null && l2!=null) return l2;
       if(l1!=null && l2==null) return l1;
       //定义头指针
       if(l1.val<=l2.val){
           //分别对l1和l2取当前和下一节点指针
           ListNode first=l1;
           ListNode second=l1.next;
           ListNode l2_temp=l2;
           ListNode l2_second=l2.next;
           //l2插向l1
           while(l2_temp!=null){
               if(second!=null){
               if(l2_temp.val>=first.val && l2_temp.val<=second.val){
                  first.next=l2_temp;
                  l2_temp.next=second;
                  first=first.next;
                  second=first.next;
                  //这个try catch是指当l2_second=null时，l2_second.next是空指针，这时已经排序结束，所以return
                 try{
                  l2_temp=l2_second;
                  l2_second=l2_second.next;}
                 catch (Exception e){
                	 return l1;
                 }

               }
               else{
            	      first=first.next;
                   second=first.next;
               }}
               else{
                   first.next=l2_temp;
                  return l1;
               }

           }
           return l1;
       }
       else{
    	   ListNode first=l2;
           ListNode second=l2.next;
           ListNode l1_temp=l1;
           ListNode l1_second=l1.next;
           while(l1_temp!=null){
               if(second!=null){
               if(l1_temp.val>=first.val && l1_temp.val<=second.val){
                  first.next=l1_temp;
                  l1_temp.next=second;
                  first=first.next;
                  second=first.next;
                 try{
                  l1_temp=l1_second;
                  l1_second=l1_second.next;}
                 catch (Exception e){
                	 return l2;
                 }

               }
               else{
            	      first=first.next;
                   second=first.next;
               }}
               else{
                   first.next=l1_temp;
                  return l2;
               }

           }
           return l2;

       }

    }

}

代码下载：https://github.com/jimenbian/GarvinLeetCode

/********************************

* 本文来自博客  “李博Garvin“

* 转载请标明出处:http://blog.csdn.net/buptgshengod

******************************************/