/*
public class ListNode {int val;ListNode next = null;ListNode(int val) {this.val = val;}
}*/
public class Solution {public ListNode Merge(ListNode list1,ListNode list2) {if(list1==null)return list2; //判断到某个链表为空就返回另一个链表。如果两个链表都为空呢？没关系，这时候随便返回哪个链表，不也是空的吗?if(list2==null)return list1;ListNode list0=null;//定义一个链表作为返回值if(list1.val<list2.val){//判断此时的值，如果list1比较小，就先把list1赋值给list0，反之亦然list0=list1;list0.next=Merge(list1.next, list2);//做递归，求链表的下一跳的值}else{list0=list2;list0.next=Merge(list1, list2.next);}return list0;}
}

简化一下，用那个三目运算符：

public class Solution {public ListNode Merge(ListNode list1,ListNode list2) {if(list1==null)return list2;if(list2==null)return list1;ListNode head;
list0= list1.val>list2.val?list2:list1;
list0.next = list1.val>list2.val?Merge(list1,list2.next):Merge(list1.next,list2);
return list0;
}
}

/*
public class ListNode {int val;ListNode next = null;ListNode(int val) {this.val = val;}
}*/
public class Solution {public ListNode Merge(ListNode list1,ListNode list2) {if(list1 == null)return list2;if(list2 == null )return list1;ListNode tmp1 = list1;ListNode tmp2 = list2;ListNode head = new ListNode(0);//这里不能把返回链表赋值为null，因为下一行马上就要把它赋值给另一链表，得让它在内存里有位置才行ListNode headptr = head;while(tmp1 != null && tmp2!=null){             if(tmp1.val <= tmp2.val){head.next=tmp1;head = head.next;tmp1 = tmp1.next;}else{head.next=tmp2;head = head.next;tmp2=tmp2.next;}}//其中一个链表已经跑到头之后，继续单链表的合并while(tmp1 != null){head.next = tmp1;head = head.next;tmp1= tmp1.next;}while(tmp2 != null){head.next = tmp2;head = head.next;tmp2= tmp2.next;}head = headptr.next;return head;}
}