面试题：递归实现两个链表合并

[napoleonpan]

两个升序链表，合并成一个升序链表，要用递归做！

不用递归如下：

Node* mergeAction(Node* head1, Node* head2){//两个链表的合并操作
	Node* head=(Node*)malloc(sizeof(Node));
	Node* q=head;
	while(head1 && head2){
		if(head1->data<=head2->data){
			Node* p=(Node*)malloc(sizeof(Node));
			p->data=head1->data;
			p->next=NULL;
			q->next=p;
			q=q->next;
			head1=head1->next;
		}
		else if(head1->data > head2->data){
			Node* p=(Node*)malloc(sizeof(Node));
			p->data=head2->data;
			p->next=NULL;
			q->next=p;
			q=q->next;
			head2=head2->next;
		}
	}
	return head->next;
}

[bitxinhai]

L1,L2,L都是无头节点的单链表，下面程序可以实现！！！
void MergeList(LinkList L1,LinkList L2,LinkList &L)
{/*两个升序链表，合并成一个升序链表，要用递归做！*/
LinkList p1 = L1;
LinkList p2 = L2;

if(p1 && p2)
{
if(p1->data < p2->data)
{
L->next = p1;
MergeList(L1->next,L2,L->next);
}
else
{
L->next = p2;
MergeList(L1,L2->next,L->next);
}

}
else if(p1)
{
L->next = p1;
}
else
if(p2)
{
L->next = p2;
}

}

[w382120035]

Node* mergAction(Node* head1,Node *head2)
{
   Node *p=NULL;
   if(head1==NULL&&head2==NULL)
	   return p;
   else if(head1==NULL)
	   return head2;
   else if(head2==NULL)
	   return head1;
   else
   {
		if(head1->data < head2->data)
		{
			p = head1;
			p->next = mergAction(head1->next,head2);
		}
		else
		{
		   p = head2;
		   p->next = mergAction(head1,head2->next);
		}
		return p;
   }
}

[林初茵]

ddddddddddddddddddddddddddd

[napoleonpan]

强人啊