一个单向的复杂链表，每个节点有两个指针，一个是next，一个是any指针。next指针指向下一个节点，any指针可以指向任意一个节点包括NULL。

把这个链表复制一遍，要求任意指针指向复制链表相对应的位置。


我的思想是先在原来的节点后面都加一个新的节点，后面的节点存原来节点的信息。

第一步，先在每个旧节点后面创建一个新节点，并把值付给新节点。

第二步，再重新遍历这个新的链表，让新节点的any指针指向前一个旧节点的any指针指向的下一个节点。

第三步，把加进去的节点一个一个断开，恢复旧的链表。并连成新的链表，此链表就是复制好的链表。


#include<stdio.h>
#include<stdlib.h>

typedef struct node
{
int num;
struct node* next;
struct node* any;
}NODE;

NODE* create(NODE* phead,int nu)
{
NODE* tmp = malloc(sizeof(struct node));
tmp->num = nu;
if(phead == NULL)
return tmp;
else{
while(phead->next != NULL)
phead = phead->next;
phead->next = tmp;
return tmp;
}
}

void show(NODE* head)
{
while(head){
printf("%d ",head->num);
if(head->any != NULL)
printf("%d ",head->any->num);
else

printf("NULL ");

head = head->next;
printf("\n");
}
}

NODE* copy(NODE* phead, NODE* head)
{
NODE* newhead;
NODE* tmp;
NODE* newtail;
NODE* head1 = head;
phead = head;
while(head){
NODE* tmp = malloc(sizeof(struct node));
tmp->num = head->num;
tmp->next = head->next;
head->next = tmp;
head = head->next->next;
}
while(phead){
if(phead->any == NULL)
phead->next->any = NULL;
else{
phead->next->any = phead->any->next;
}
phead = phead->next->next;
}
newhead = head1->next;
newtail = newhead;
head1 = head1->next->next;
while(head1){
tmp = head1->next;
head1->next = head1->next->next;
tmp->next = NULL;
newtail->next = tmp;
newtail = tmp;
head1 = head1->next;
}
return newhead;
}

int main()
{
NODE* head = NULL;
NODE* phead = NULL;
NODE* tail = head;
head = create(head,20);
head->any = NULL;
tail = create(head,21);
tail->any = NULL;
tail = create(tail,22);
tail->any = head;
tail = create(tail,23);
tail->any = head->next;
tail = create(tail,24);
tail->any = head->next->next;
tail = create(tail,25);
tail->any = NULL;
tail = create(tail,26);
tail->any = head;
show(head);
printf("the copy is: \n");
phead = copy(phead,head);
show(phead);
return 0;
}