/* 
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; 
  } 
} 

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;
 }
}