一，题目
两个非降序链表的并集，1->2->3
和 2->3->5 并为 1->2->3->5。另外只能输出结果，不能修改两个链表的数据。
二，递归解法
#include <iostream>
using namespace std;
struct Node
{
int data;
Node * next;
};
Node * MergeRecursive(Node *head1 , Node *head2)
{
if ( head1 == NULL )
return head2 ;
if ( head2 == NULL)
return head1 ;
Node *head = new Node() ;
if ( head1->data < head2->data )
{
head = head1 ;
head->next = MergeRecursive(head1->next,head2);
}
else
{
head = head2 ;
head->next = MergeRecursive(head1,head2->next);
}
return head ;
}
Node *creatLink(int a[],int n)
{
Node *head=new Node();
head->data=a[0];
Node *tail=head;
head->next=NULL;
for(int i=1;i<n;++i)
{
Node *temp=new Node();
temp->data=a[i];
temp->next=NULL;
tail->next=temp;
tail=tail->next;
}
return head;
}
int main()
{
int a[]={1,2,4,6,7};
int b[]={2,3,5,7,8};
Node *headA=creatLink(a,5);
Node *headB=creatLink(b,5);
Node *headResult= MergeRecursive(headA , headB);
while(headResult)
{
cout<<headResult->data<<" ";
headResult=headResult->next;
}
}