环链表转单链表

    昨儿晚去笔了家公司，然后看到一个环链表转单链表的题。题是这样的，有个环链表，依次把第m个节点删除环链表（约瑟夫环操作）。最后根据节点删除顺序重建为单向链表，并返回。

void reorder(node **head, int m)
{
    // Check param.
    if (head == NULL || *head == NULL || m <= 0)
    {
        return;
    }

    // Special case.
    if (m == 1)
    {
        node *p = *head;
        while (p->next != *head) p = p->next;
        p->next = NULL;

        return;
    }

    node **backup = head;
    node *iter = *head, *curr = NULL, *ret = NULL;
    int n = 0;

    while (iter->next != iter)
    {
        ++n;

        // iter always equals *backup.
        if (!(n % m))
        {
            if (ret == NULL)
            {
                ret = curr = iter;
            }
            else
            {
                curr->next = iter;
                curr = curr->next;
            }

            // Update backup.
            *backup = (*backup)->next;
        }
        else
        {
            // Update backup.
            backup = &iter->next;
        }

        iter = iter->next;
    }

    // Link last node.
    if (iter != curr)
    {
        curr->next = iter;
        iter->next = NULL;
    }

    // Write return value.
    *head = ret;
}