【链表】15、判断循环双链表是否对称

// createDouLoopLink.cpp
#define _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>
#include <stdio.h>
struct Link {
    int data;
    struct Link *next;
    struct Link *pre;
};
Link *createDouLoopLink() {
    int n, data;
    struct Link *head = (struct Link *)malloc(sizeof(struct Link));
    head->next = NULL;
    head->pre = NULL;
    struct Link *p = head;
    printf("请输入节点个数：n=");
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        printf("请输入第%d个节点值:", i + 1);
        scanf("%d", &data);
        struct Link *newP = (struct Link*)malloc(sizeof(struct Link));
        newP->data = data;
        newP->pre = p;
        p->next = newP;
        p = newP;
    }
    p->next = head;
    head->pre = p;
    return head;
}

/*
    设计一个算法判断带头结点的循环双链表是否对称
    分析：
        简单分析，我们可以设置两个指针，pre和next，从头结点出发，进行比较，若pre与next所指值不同，则不对称，若pre和next指向了同一个节点
        则该循环双链表对称
*/
struct Link {
    int data;
    struct Link *next;
    struct Link *pre;
};
#define _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>
#include <stdio.h>
void isSymmetry(Link *h) {
    struct Link *pre = h->pre,*next=h->next;
    while (pre!=next&&pre->pre!=next) {//此时存在两种情况，奇数个节点和偶数个节点都要考虑
        if (pre->data!=next->data) {
            printf("该循环双链表不对称");
            break;
        }
        else {
            pre = pre->pre;
            next = next->next;
        }
    }
    if (pre==next||pre->pre==next) {
        printf("该循环双链表对称");
    }
}
int main() {
    struct Link *head;
    Link *createDouLoopLink();
    head = createDouLoopLink();
    isSymmetry(head);
    return 0;
}