调试归并

Question

我需要排序双向链表。根据全能维基百科，归并是去该方式。

在递归算法的性能非常好，但我正在写一个通用的实现，性能可能是一个问题。

移植数组迭代版本将杀死性能重新扫描列表以将其分成子列表缓慢;对于任何有兴趣 - 这里的代码：

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;
        }
    }
}

但是有使用基于堆栈的方法的另一迭代版本：

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);
}

这将推动大小1所列出入堆栈，只要顶端列表比其前身大于或等于大小的合并了下来。

不幸的是，有一个错误的地方，作为一些输入列表，merge_down()将无法进行仔细的检查：

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;
        }
    }
}

在链表实现可能是相关的，以及：

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;
}

我是什么在这里失踪？

Answer 1

我发现了错误自己：

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);
}

下面，current的下一个值被确定的之后调用merge_down()，这可能会左右移动节点，即current->next将不再指向正确的节点。

重新排列解决了这个问题：

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

由于PMG的努力：我添加了一些票为

Answer 2

你是否需要一个节点复制到列表的末尾？点击 什么是insert_before()通话呢？

insert_before(node, list, NULL);

这会弄乱list->first和node->prev。

Answer 3

我现在已经运行你的代码，并得到了后，我注释掉线下方所示的工作。

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; /* possible bug? */
 /* ^^^^^^^^^^^^^^^^^^^ */

这是否对你的工作吗？

Answer 4

基于堆栈的方法

/* ... */
    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);
    }
/* ... */

top总是将成为0的第一次循环时，正确？结果 该merge_down()功能是永远不会被调用。我没有尝试的代码，但不看的权利。

---

修改的结果， 为stack 32个元件是不够...当列表具有为了多于32个元件（几个道次之后也许）你写超出stack的末尾。

Answer 5

作为KRISS问它，这里的递归版本（一个标准的归并排序使用合并函数从其他exaples）：

static struct node *merge(struct linked_list *list,
    int (*cmp)(const void *, const void *),
    struct node *left, struct node *right, size_t right_count)
{
    struct node *head;
    if(cmp(left->data, right->data) <= 0)
    {
        head = left;
        left = left->next;
    }
    else
    {
        head = right;
        struct node *node = right;
        right = right->next;
        remove_node(node, list);
        insert_before(node, list, left);
        --right_count;
    }

    while(left != right && 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);
            --right_count;
        }
    }

    return head;
}

static struct node *mergesort(struct linked_list *list,
    int (*cmp)(const void *, const void *), struct node *head, size_t size)
{
    if(size < 2) return head;
    size_t left_count = size / 2;
    size_t right_count = size - left_count;

    struct node *tail = head;
    size_t count = left_count;
    while(count--) tail = tail->next;

    return merge(list, cmp,
        mergesort(list, cmp, head, left_count),
        mergesort(list, cmp, tail, right_count),
        right_count);
}

static void sort(struct linked_list *list,
    int (*cmp)(const void *, const void *))
{
    mergesort(list, cmp, list->first, list->size);
}