c - 调试合并排序

Question

我需要对一个双向链表进行排序。根据全能的维基百科，合并排序是实现这一目标的方法。

递归算法工作得相当好，但由于我正在编写通用实现，性能可能是个问题。

移植数组的迭代版本会降低性能，因为重新扫描列表以将其划分为子列表很慢；对于任何感兴趣的人 - 这是代码:

static void sort(struct linked_list *list,
    int (*cmp)(const void *, const void *))
{
    size_t slice_size = 1;
    for(; slice_size < list->size; slice_size *= 2)
    {
        struct node *tail = list->first;
        while(tail)
        {
            struct node *head = tail;

            size_t count = slice_size;
            while(tail && count--) // performance killer
                tail = tail->next;

            count = slice_size;
            while(head != tail && tail && count)
            {
                if(cmp(head->data, tail->data) <= 0)
                    head = head->next;
                else
                {
                    struct node *node = tail;
                    tail = tail->next;
                    remove_node(node, list);
                    insert_before(node, list, head);
                    --count;
                }
            }

            while(tail && count--) // performance killer
                tail = tail->next;
        }
    }
}

但是还有另一个使用基于堆栈的方法的迭代版本:

struct slice
{
    struct node *head;
    size_t size;
};

static void sort(struct linked_list *list,
    int (*cmp)(const void *, const void *))
{
    if(list->size < 2) return;

    struct slice stack[32];
    size_t top = -1;
    struct node *current = list->first;

    for(; current; current = current->next)
    {
        stack[++top] = (struct slice){ current, 1 };
        for(; top && stack[top-1].size <= stack[top].size; --top)
            merge_down(list, cmp, stack + top);
    }

    for(; top; --top)
        merge_down(list, cmp, stack + top);
}

这会将大小为 1 的列表压入堆栈并向下合并，只要顶部列表的大小大于或等于其前身。

不幸的是，对于某些输入列表，某处存在错误，merge_down() 将无法通过健全性检查:

static void merge_down(struct linked_list *list,
    int (*cmp)(const void *, const void *), struct slice *top)
{
    struct node *right = top->head;
    size_t count = top->size;

    --top;

    struct node *left = top->head;
    top->size += count;

{
    // sanity check: count nodes in right list
    int i = count;
    struct node *node = right;
    for(; i--; node = node->next) if(!node)
    {
        puts("too few right nodes");
        exit(0);
    }
}

    // determine merged list's head
    if(cmp(left->data, right->data) <= 0)
    {
        top->head = left;
        left = left->next;
    }
    else
    {
        top->head = right;
        struct node *node = right;
        right = right->next;
        remove_node(node, list);
        insert_before(node, list, left);
        --count;
    }

    while(left != right && count)
    {
        if(cmp(left->data, right->data) <= 0)
            left = left->next;
        else
        {
            struct node *node = right;
            right = right->next;
            remove_node(node, list);
            insert_before(node, list, left);
            --count;
        }
    }
}

链表实现也可能相关:

struct node
{
    struct node *prev;
    struct node *next;
    long long data[]; // use `long long` for alignment
};

struct linked_list
{
    struct _list _list; // ignore
    size_t size;
    struct node *first;
    struct node *last;
};

static void insert_before(struct node *node, struct linked_list *list,
    struct node *ref_node)
{
    if(ref_node)
    {
        node->next = ref_node;
        node->prev = ref_node->prev;
        if(ref_node->prev) ref_node->prev->next = node;
        else list->first = node;
        ref_node->prev = node;
    }
    else // empty list
    {
        node->next = NULL;
        node->prev = NULL;
        list->first = node;
        list->last = node;
    }
    ++list->size;
}

static void remove_node(struct node *node, struct linked_list *list)
{
    if(node->prev) node->prev->next = node->next;
    else list->first = node->next;
    if(node->next) node->next->prev = node->prev;
    else list->last = node->prev;
    --list->size;
}

我在这里错过了什么？

Answer 1

你是否需要将节点复制到列表的末尾？
那么 insert_before() 调用是什么？

insert_before(node, list, NULL);

那会弄乱 list->first 和 node->prev。