排序双链表C

尝试通过遍历列表的循环来实现.

在循环中,我将头节点输入到已定义的排序函数中,然后使用strcmp来确定节点中的哪个名称应优先出现.

它不起作用,因为过早写入名称.

我通过一次向下列出一个节点,而不回头看看第一个节点是否应该在最后一个节点之前,对它们进行线性比较.解释的那部分会有所帮助.

现在,对我来说最重要的两个功能定义如下：
我已尽力去做我认为适合排序功能的事情.

void list::displayByName(ostream& out) const
{
    list *ListPtr = NULL;
    node *current_node = headByName;
    winery *wine_t = new winery(); 
    // winery is another class object type
    // im allocating it to prevent a crash when I call it.

    while ( current_node != NULL )
    {
        *(wine_t) = current_node->item;
        wine_t = ListPtr->sort( current_node );
        out << wine_t << endl;
        current_node = current_node->nextByName;
    }
    delete wine_t;
}

winery * const list::sort( node * current_node ) const
{
    // current_node is the first node.
    const char *SecondName = NULL, *FirstName  = NULL;
    winery *wine_t   = new winery();

    if ( current_node != NULL )
    {
        SecondName   = current_node->item.getName();            
        current_node = current_node->nextByName;
        FirstName    = current_node->item.getName();
    }

    if ( strcmp( FirstName, SecondName ) == -1 )
    {
        *(wine_t)  = current_node->item;
        FirstName  = NULL;
        SecondName = NULL;
        return wine_t;
    }
    else if ( strcmp( FirstName, SecondName ) == 1 )
    {
        *(wine_t) = current_node->item;
        FirstName = NULL;
        SecondName = NULL;
        return wine_t;
    }
    else return wine_t;// then the strings are equal

    FirstName  = NULL;
    SecondName = NULL;
    return wine_t;
}

我开始在这里开发节点：

void list::insert(const winery& winery)
{
    node *current_node = new node( winery );

    if ( headByName == NULL )
    {
        headByName   = current_node;
        headByRating = current_node;
        tail         = headByName;
        current_node->prev = current_node;
    }
    else
    {
        current_node->prev = tail;
        tail->nextByName   = current_node;
    }

    tail = current_node;
    current_node = NULL;
}

我认为在上述功能中它是正确的.我可以在那里整理一下吗？

以下是我正在使用的变量：

public list
{
         ...
    void insert(const winery& winery);
    void displayByName(ostream& out) const;
}
private:
    struct node
    {
        node(const winery& winery);     // constructor
        winery item;
        node * prev;
        node * nextByName;
        node * nextByRating;
    };

    winery * const sort(node*) const;
    node * headByName;
    node * headByRating;
    node * tail;
};

任何帮助表示赞赏.
非常感谢=)

解决方法:

据我了解,您希望list :: sort在列表中找到大于输入的最小节点.

为此,您需要遍历所有元素,并保持找到当前最小但更大的节点.

像这样：

node * const list::sort( node * given_node ) const
{
    if ( given_node == NULL )
    {
        return NULL;
    }

    // Smallest node found which is greater than given_node.
    node * least_found_node = NULL;

    // Node we are looking at right now.
    node * current_node = given_node->nextByName;

    // Go through all nodes.
    while ( current_node && current_node != given_node )
    {
        // Is this node bigger than the given node?
        if ( strcmp( current_node->item.getName(), given_node->item.getName() ) < 0 )
        {
            // Is this node smaller than the smallest node we know of?
            if ( least_found_node == NULL ||
               ((strcmp( current_node->item.getName(), least_found_node->item.getName() ) > 0) )
            {
                // We found a better node.
                least_found_node = current_node;
            }
        }

        current_node = current_node->nextByName;
    }

    return least_found_node;
}

现在,更改显示功能以使用它,如下所示：

void list::displayByName(ostream& out) const
{
    // Find first node initially.
    node * current_node = sort( NULL );

    while ( current_node != NULL )
    {
        // Print node.
        out << current_node->item.getName();

        // Find next node in sorted output.
        current_node = sort( current_node );
    }
}

这部分将继续调用sort,直到sort返回NULL.排序的第一个调用是NULL,因此找到了最低的项目(即排序列表中的第一个).如果没有更多的节点大于current_node,则sort返回NULL,从而终止循环.

标签：c,sorting,linked-list
来源： https://codeday.me/bug/20191009/1881994.html