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;
}
}
}
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* 本文来自博客 “李博Garvin“
* 转载请标明出处:http://blog.csdn.net/buptgshengod
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