《剑指offer》面试题17—合并两个排序链表

题目：输入两个递增排顺序的链表，合并这两个链表并使合并后的链表仍是递增排序的。

重点：

1.代码鲁棒性：要判断输入的两个链表都为NULL；其中一个链表为NULL的情况。

2.用递归实现，注意递归的思路。

 #include <iostream>

 using namespace std;

 struct Node
 {
     int val;
     Node* next;
 };

 void PrintLink(Node* phead)
 {
     Node* p = phead;
     while(p != NULL)
     {
         int temp = p->val;
         cout<<temp<<endl;
         p = p->next;
     }
 }

 Node* Merge(Node* phead1, Node* phead2)
 {
     if(phead1 == NULL && phead2 == NULL) return NULL;
     else if(phead1 == NULL && phead2 != NULL) return phead2;
     else if(phead1 != NULL && phead2 ==NULL) return phead1;
     else
     {
         Node* pselected;
         if(phead1->val < phead2->val)
         {
             pselected = phead1;
             pselected->next = Merge(phead1->next,phead2);
         }
         else
         {
             pselected = phead2;
             pselected->next = Merge(phead1,phead2->next);
         }
     return pselected;
     }

 }

 int main()
 {
     int n;
    // while(scanf("%d",&n)!=EOF)
   cin>>n;
         Node * head1;
         if(!n)
         head1=NULL;
         else
         head1=new Node();
         Node * p=head1;
         for(int i=;i<n;++i)
         {
             scanf("%d",&p->val);
             if(i==n-)
             {
                 p->next=NULL;
                 break;
             }
             p->next=new Node();
             p=p->next;
         }
         PrintLink(head1);

         scanf("%d",&n);
         Node * head2;
         if(!n)
         head2=NULL;
         else
         head2=new Node();
         p=head2;
         for(int i=;i<n;++i)
         {
             scanf("%d",&p->val);
             if(i==n-)
             {
                 p->next=NULL;
                 break;
             }
             p->next=new Node();
             p=p->next;
         }
         PrintLink(head2);
         Node * newhead=Merge(head1,head2);
         PrintLink(newhead);
     return ;
 }