/* 
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; //判斷到某個鏈表為空就返回另一個鏈表。如果兩個鏈表都為空呢？沒關系，這時候隨便返回哪個鏈表，不也是空的嗎? 
    if(list2==null)return list1; 
    ListNode list0=null;//定義一個鏈表作為返回值 
    if(list1.val<list2.val){//判斷此時的值，如果list1比較小，就先把list1賦值給list0，反之亦然 
      list0=list1; 
      list0.next=Merge(list1.next, list2);//做遞歸，求鏈表的下一跳的值 
      } 
      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);
		//這里不能把返回鏈表賦值為null，因為下一行馬上就要把它賦值給另一鏈表，得讓它在內存里有位置才行 
		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;
			}
		}
		//其中一個鏈表已經跑到頭之后，繼續(xù)單鏈表的合并 
		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;
	}
}