《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=0;i<n;++i)
        {
            scanf("%d",&p->val);
            if(i==n-1)
            {
                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=0;i<n;++i)
        {
            scanf("%d",&p->val);
            if(i==n-1)
            {
                p->next=NULL;
                break;
            }
            p->next=new Node();
            p=p->next;
        }
        PrintLink(head2);
        Node * newhead=Merge(head1,head2);
        PrintLink(newhead);
    return 0;
}