设以带头结点的双向循环链表表示的线性表L= (a1,a2,…,an),试写一时间复杂度O(n)的算法，将L改造为 (a1,a3,…,an,…,a4,a2)。

#include<stdio.h>
typedef int ElemType;
typedef struct DuLNode{
ElemType data;
struct DuLNode *prior;
struct DuLNode *next;
} DuLNode, *DuLinkList;

DuLinkList InitDuLinkList();//初始化双向循环链表(头节点)
void CreateDuLinkList(DuLinkList L, ElemType *arr, int n);//创建链表元素
void ShowList(DuLinkList L);//输出链表

/*
将L中的元素，按如下规则插入新表，并返回新表。
（1,2）->(1,3,2)->(1,3,4,2)->(1,3,5,4,2)->(1,3,5,6,4,2)->...
*/
DuLinkList Transform(DuLinkList L);

int main()
{
ElemType data[7] = { 1, 2, 3, 4, 5, 6, 7 };//测试数据1（奇数个）
//ElemType data[8] = { 1, 2, 3, 4, 5, 6, 7, 8 };//测试数据2（偶数个）
int length = sizeof(data) / sizeof(ElemType);
DuLinkList L = InitDuLinkList(), Lnew;
CreateDuLinkList(L, data, length);
printf("原链表：");
ShowList(L);

Lnew = Transform(L);
printf("改造表：");
ShowList(Lnew);

getchar();
return 0;
}

DuLinkList InitDuLinkList()
{
DuLNode *dnode = (DuLNode *)malloc(sizeof(DuLNode));
dnode->data = 0;
dnode->prior = dnode;
dnode->next = dnode;
return dnode;
}

void CreateDuLinkList(DuLinkList L, ElemType *arr, int n)
{
DuLNode *dnode;
DuLNode *p = L;
int i;
for (i = 0; i < n; i++)
{
dnode = (DuLNode *)malloc(sizeof(DuLNode));
dnode->data = arr[i];
dnode->next = L;
dnode->prior = p;
p->next = dnode;
p = p->next;
}
}

void ShowList(DuLinkList L)
{
int i;
DuLNode *r = L->next;
while (r->next != L){
printf("%d ", r->data);
r = r->next;
}
printf("%d ", r->data);
printf("\n");
}

DuLinkList Transform(DuLinkList L)
{
DuLinkList Lnew = InitDuLinkList();
DuLNode *p, *q, *pa, *pb;
q = p = L->next;
pa = pb = Lnew;

while (p != L)
{
if (p != L)
{
/*L中寄数个数据插入Lnew*/
q = p->next;//保留 L 链表
//pa之后插入p
p->prior = pa;
pa->next = p;
p->next = pb;
pb->prior = p;

pa = pa->next;
p = q;//p指向 待操作 L
}
if (p != L)
{
/*L中偶数个数据插入Lnew*/
q = p->next;//保留 L 链表
//pb之前插入p
p->next = pb;
pb->prior = p;
p->prior = pa;
pa->next = p;

pb = pb->prior;
p = q;//p指向 待操作 L
}
}
return Lnew;
}