已知两个链表A和B分别表示两个集合, 其元素递增排列. 设计一个算法, 求出A与B的交集, 并存放在A链表中

我不是女神ヾ 2022-10-15 00:42 7阅读 0赞

#include<stdio.h>
    #include<stdlib.h>
    short InitHeadNode(struct Node* &, struct Node* &);
    short RearInsert(struct Node* &, struct Node* &, short);
    short GetListLength(struct Node* &);
    short IsElemIn(struct Node* &, short);
    void DeleteNodeNotIn1ButIn2(struct Node* &, struct Node* &);
    void Traversal(struct Node* &);
    /* 链表结点定义 */
    struct Node
    {
        short data;
        struct Node *next;
    };
    int main()
    {
        struct Node *head1;
        struct Node *rear1;
        struct Node *head2;
        struct Node *rear2;
        /* 初始化1#链表的头结点 */
        if(InitHeadNode(head1, rear1) == -1)
        {
            /* 头结点申请失败, 程序直接结束 */
            exit(0);
        }
        short temp;
        /* 向链表中插入数据结点 */
        while(1)
        {
            while(scanf("%hd", &temp) != 1)
            {
                while(getchar() != '\n') ;
                printf("请输入合法数据.\n");
            }
            if(temp == -1)
            {
                break;
            }
            if(RearInsert(head1, rear1, temp) == -1)
            {
                /* 申请结点失败 */
                exit(0);
            }
        }
        /* 初始化2#链表的头结点 */
        if(InitHeadNode(head2, rear2) == -1)
        {
            /* 头结点申请失败, 程序直接结束 */
            exit(0);
        }
        /* 向链表中插入数据结点 */
        while(1)
        {
            while(scanf("%hd", &temp) != 1)
            {
                while(getchar() != '\n') ;
                printf("请输入合法数据.\n");
            }
            if(temp == -1)
            {
                break;
            }
            if(RearInsert(head2, rear2, temp) == -1)
            {
                /* 申请结点失败 */
                exit(0);
            }
        }
        /* 至此, A链表和B链表均已初始化 */
        DeleteNodeNotIn1ButIn2(head2, head1);
        Traversal(head1);
        /* */
        return 0;
    }
    /* 判断数据元素value是否在链表中 */
    short IsElemIn(struct Node* &head, short value)
    {
        struct Node *p = head -> next;
        while(p != NULL)
        {
            if(p -> data == value)
            {
                /* 找到数据元素value */
                return 1;
            }
            p = p -> next;
        }
        /* 未找到数据元素value */
        return -1;
    }
    /* 删除以下2#集合中的元素结点: 在2#集合中出现但未在1#集合中出现的元素结点 */
    void DeleteNodeNotIn1ButIn2(struct Node* &head1, struct Node* &head2)
    {
        struct Node *p2 = head2 -> next;
        struct Node *q2 = head2;
        /* 删除在集合2中出现但未在集合1中出现的结点 */
        while(p2 != NULL)
        {
            if(IsElemIn(head1, p2 -> data) == -1)
            {
                /* p2所指数据元素(集合2中)在集合1中未出现 */
                /* 删除该结点 */
                q2 -> next = p2 -> next;
                free(p2);
                p2 = q2 -> next;
            }
            else
            {
                p2 = p2 -> next;
                q2 = q2 -> next;
            }
        }
    }
    /* 遍历单链表并输出 */
    void Traversal(struct Node* &head)
    {
        struct Node *p = head -> next;
        while(p != NULL)
        {
            printf("%3hd", p -> data);
            p = p -> next;
        }
        putchar('\n');
    }
    /* 获取表长 */
    short GetListLength(struct Node* &head)
    {
        struct Node *p = head -> next;
        short length = 0;
        while(p != NULL)
        {
            length ++;
            p = p -> next;
        }
        return length;
    }
    /* 尾插法 */
    short RearInsert(struct Node* &head, struct Node* &rear, short value)
    {
        struct Node *p = (struct Node*)malloc(sizeof(struct Node));
        if(p != NULL)
        {
            /* 结点申请成功 */
            rear -> next = p;
            p -> next = NULL;
            p -> data = value;
            rear = p;
            return 1;
        }
        else
        {
            /* 结点申请失败 */
            return -1;
        }
    }
    /* 初始化头结点 */
    short InitHeadNode(struct Node* &head, struct Node* &rear)
    {
        struct Node *p = (struct Node*)malloc(sizeof(struct Node));
        if(p != NULL)
        {
            /* 头结点申请成功 */
            head = p;
            head -> data = 0;
            head -> next = NULL;
            rear = head;
            return 1;
        }
        else
        {
            /* 头结点申请失败 */
            return -1;
        }
    }

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[20210522201417662.png]: /images/20221014/66d322cd7e79407daccfc84557994234.png
[20210522201502426.png]: /images/20221014/9bb96f8cdaa74c44ab3d0136e83d1fc0.png
[20210522201611824.png]: /images/20221014/b88ae40b057442e89d9637da666b51b6.png