Question

我正在尝试计算链表中IP的出现次数，并打印出每个IP地址的计数。我可以打印计数但是当我迭代它时，计数减少并且重复打印。此外，它们应按最高登录次数排序。

我尝试使用数组实现此操作，方法是为每个IP创建数组副本，但这样做无效，因为无法事先为每个用户确定数组的大小。

预期产出：

    IP Address Count
    --------------------
    97.88.145.98 12
    141.219.153.201 4
    141.219.209.223 3
    141.219.210.114 3
    141.219.208.123 1
    75.129.96.98 1
    141.219.210.180 1

当前输出：

  IP Address    Occurence
---------------------------
    141.219.153.201     4
    141.219.153.201     3
    97.88.145.98        12
    97.88.145.98        11
    97.88.145.98        10
    97.88.145.98        9
    97.88.145.98        8
    141.219.210.114     3
    141.219.210.114     2
    141.219.210.114     1
    97.88.145.98        7
    97.88.145.98        6
    141.219.210.180     1
    97.88.145.98        5
    97.88.145.98        4
    97.88.145.98        3
    97.88.145.98        2
    75.129.96.98        1
    141.219.209.223     3
    141.219.209.223     2
    141.219.209.223     1
    97.88.145.98        1
    141.219.153.201     2
    141.219.208.123     1
    141.219.153.201     1

迭代链表的函数。

bstNode* search(char* key, bstNode* root)
{
    int res;
    bstNode *leaf = root;

    if( leaf != NULL ) {
        res = strcmp(key, leaf->data);
        if( res < 0)
            search( key, leaf->left);
        else if( res > 0)
            search( key, leaf->right);
        else
        {
            printf("\n'%s' found!\n", key);
            printf("---------------------------\n  IP Address\tOccurence\n---------------------------\n");
            //int count = 0;
            IP *temp = leaf->ipHead;

            while (temp) {

                int tempip = temp->ip;
                int ipcount = 0;



                uint32_t ip = tempip;
                struct in_addr ip_addr;
                ip_addr.s_addr = ip;

                //bstNode *cpy = leaf;

                ipcount = count(temp, tempip);
                //temp->count = ipcount;

                //temp = leaf;

                //printf("The IP address is %s\n C:%d\n", inet_ntoa(ip_addr), ipcount);
                printf("    %s\t\t%i\n", inet_ntoa(ip_addr), ipcount);
                temp = temp->ipNext;
            }
        }
    }

    else printf("\nNot in tree\n");
    return leaf;
}

计算出现次数的函数：

 int count(IP* start, int item)
    {
        IP* current = start;

        int c = 0;
        while (current)
        {
            if (current->ip == item)
            {
                c++;
            }
            current = current->ipNext;
        }
        //printf("Count is: %d", c);
        return c;
    }

任何帮助都会非常感激！

Answer 1

没有任何条件声明可以在函数search（）中过滤重复的IP，更准确地说是在（＆temp）＆＃39; while（temp）＆＃39;环。

count（）只计入列表中。以及只选择与参数匹配的ip的条件。

Answer 2

您的函数正在计算列表其余部分中ip的出现次数。所以当你的ip-list await看起来像

时

temp

然后循环迭代

    ( ip1, ip2, ip1, ip2 )

要解决此问题，您必须跟踪每个已经计入的ip：错了：

    count( ip1, ( ip1, ip2, ip1, ip2 )); // => "ip1 2"
    count( ip2, ( ip2, ip1, ip2 ));      // => "ip2 2" 
    count( ip1, ( ip1, ip2 ));           // => "ip1 1"
    count( ip2, ( ip2 ));                // => "ip2 1"

右：

    ipcount = count(temp, tempip);
    printf(...);

当然，必须首先清空簿记数据。

如何在链接列表中打印出重复次数？

2 个答案: