Question

我尝试编写一个函数来从GList *list未排序的链表中删除重复项，并返回GList *list而不重复：

GList *remove_dup (GList *list)
{
    GList *a, *b, *dup;
    a = list;

    /* Pick elements one by one */
    while (a != NULL && a->next != NULL) {
        b = a;

        /* Compare the picked element with rest of the elements */
        while (b->next != NULL) {
            /* If duplicate then delete it */
            if (a->data == b->next->data) {
                /* sequence of steps is important here */
                dup = b->next;
                b->next = b->next->next;
                g_list_free_1 (dup);
            } else  /* This is tricky */ {
                b = b->next;
            }
        }
        a = a->next;
    }

    /* return list without duplicates */
    return list;
}

示例name.list带有重复项：

A
A
B
C
B
A

使用后remove_dup函数：

name.list = remove_dup (name.list);

name.list没有重复：

A
B
C

- ＆GT;但似乎remove_dup返回相同的name.list重复项。

这是我在这段代码中的错误？

Answer 1

这个怎么样（没有测试过，但应该有效）：

GList *remove_dup (GList *list)
{
    GList *a, *b, *dup;

    /* Pick elements one by one */
    for (a = list; a; a = a->next) {
        /* Compare the picked element with rest of the elements */
        for (b = a->next; b;) {
            dub = b;
            b = b->next;
            if (a->data = dub->data)
                list = g_list_delete_link (list, dub);
        }
    }

    /* return list without duplicates */
    return list;
}

虽然这个O（n2）函数对于短列表来说很好，但是对于较长的列表来说效率很低。也许你想使用不同的算法，如：

GList *remove_dup (GList *list)
{
    GList *a, *dup;
    GHashTable *set;

    set = g_hash_table_new(NULL, NULL);

    /* Pick elements one by one */
    for (a = list; a;) {
        dub = a;
        a = a->next;
        if (g_hash_table_contains(set, dub->data)) {
            list = g_list_delete_link (list, dub);
        } else {
            g_hash_table_add (set, dub->data);
        }
    }

    g_hash_table_unref (set);

    /* return list without duplicates */
    return list;
}

用于从未排序的链接列表中删除重复项的功能

1 个答案: