本题代码如下

int symmetry(linklist* L)//判断循环双链表是否对称
{
    lnode* p = (*L)->next, * q = (*L)->prior;
    while (p != q && q->next != p)
    {
        if (p->data != q->data)
            return 0;
        else
        {
            p = p->next;
            q = q->prior;
        }
    }
    return 1;
}

完整测试代码

#include<stdio.h>
#include<stdlib.h>
typedef struct lnode
{
    char data;
    struct lnode* prior;
    struct lnode* next;
}lnode, * linklist;
int n = 4;
char a[4] = { 'A','B','A','B' };
void buildlinklist(linklist* L)//建立循环双链表
{
    (*L)->next = *L;//初始化头结点
    (*L)->prior = *L;
    lnode* r = *L;
    int i = 0;
    for (i = 0; i < n; i++)
    {
        lnode* s = (lnode*)malloc(sizeof(lnode));//创建新结点
        s->data = a[i];
        s->next = r->next;//插入新结点
        s->prior = r;
        r->next->prior = s;
        r->next = s;
        r = s;
    }
}
int symmetry(linklist* L)//判断循环双链表是否对称
{
    lnode* p = (*L)->next, * q = (*L)->prior;
    while (p != q && q->next != p)
    {
        if (p->data != q->data)
            return 0;
        else
        {
            p = p->next;
            q = q->prior;
        }
    }
    return 1;
}
void print(linklist* L)//输出链表
{
    lnode* k = (*L)->next;
    while (k != *L)
    {
        printf("%c ", k->data);
        k = k->next;
    }
}
int main()
{
    linklist L = (lnode*)malloc(sizeof(lnode));//创建头结点
    L->next = L;
    L->prior = L;
    buildlinklist(&L);//构建循环双链表
    printf("原始单链表为：");
    print(&L);
    int ret = symmetry(&L);
    if (ret == 1)
    {
        printf("\n带头结点的循环双链表对称");
    }
    else
    {
        printf("\n带头结点的循环双链表不对称");
    }
    return 0;
}

用char a[4] = { 'A','B','B','A' }测试

用char a[4] = { 'A','B','A','B' }测试