Question

我正在尝试将链接列表程序设置为双向链接列表，但是，当我尝试向后打印列表时遇到问题。在我尝试向后打印的那一刻，它只是运行一个永无止境的循环，我无法弄清楚错误的位置。如果有人能指出我做错了什么，那就非常有帮助。

编辑：我认为问题出在displaybackwards方法中，但我不知道如何更改它，因为删除它会导致程序崩溃。

这些是我当前代码的一部分，我认为问题可能在于：

struct  NODE
{
    union
    {
        int  nodeCounter;
        void  *dataitem;
    }item;
    struct NODE *link;
    struct NODE *backlink;
};

struct NODE *InitList()
{
    struct NODE *temp = (struct NODE*)malloc(sizeof NODE);

    temp->item.nodeCounter = 0;
    temp->link = NULL;
    temp->backlink = NULL;
    return temp;
}

void  Add2List(struct NODE *start, struct NODE *NewNode)
{
    struct NODE *current = start;

    while (current->link != NULL)
    {
        current->backlink = current; //problem should be this line
        current = current->link;
    }
    current->link = NewNode;
    NewNode->link = NULL;
    NewNode->backlink = current;

    start->item.nodeCounter++;
}

void DisplayList(struct NODE *start)
{
    struct NODE *current = start->link;

    while (current != NULL)
    {
        DisplayNode((struct inventory *)current->item.dataitem);
        current = current->link;

    }
}

void DisplayBackwards(struct NODE *start)
{
    struct NODE *current = start->link;
    while(current != NULL && current->link != NULL) //goes until current == last node
    {

        current = current->link;
        current->backlink = current;
    }

    //when current == last node
    while(current != start)// && current->backlink != NULL)
    {
        DisplayNode((struct inventory*)current->item.dataitem);
        current->link = current;
        current = current->backlink;
    }
}

Answer 1

void DisplayBackwards(struct NODE *start)
{
    struct NODE *current = start;  //current points to first node
    if(current==NULL)  //if empty list, return 
       return;

    //now we are sure that atleast one node exists
    while(current->link != NULL) //goes until current == last node
    {
        current = current->link; //keep on going forward till end of list
    }

    //start from last node and keep going back till you cross the first node
    while(current != NULL)
    {
        DisplayNode((struct inventory*)current->item.dataitem);
        current = current->backlink; //go one node back
    }
}

c双向链表打印向后

1 个答案: