交换双链表 [英] swap in doubly linked list

问题描述

我正在尝试在双向链表中交换两个节点。以下是具有交换功能的程序部分。

I am trying to swap two nodes in a doubly linked list. Below is the part of program having swap function.

 int swap (int x, int y)
{
    struct node *temp = NULL ;
    struct node *ptr1, *ptr2;
    temp = (struct node *)malloc(sizeof(struct node));


    if (head == NULL )
    {
        printf("Null Nodes");

    }
    else
    {
        ptr1 = ptr2 = head;

        int count = 1;
        while (count != x)
        {
            ptr1 = ptr1->next;
            count++;

        }



        int count2 = 1;
        while (count2 != y)
        {
            ptr2 = ptr2->next;
            count2++;   

        }       


        ptr1->next->prev = ptr2;
        ptr1->prev->next = ptr2;
        ptr2->next->prev = ptr1;
        ptr2->prev->next = ptr1;

    temp->prev = ptr1->prev;
    ptr1->prev = ptr2->prev;
    ptr2->prev = temp->prev;

    temp->next = ptr1->next;
    ptr1->next = ptr2->next;
    ptr2->next = temp->next;


    }
    return 0;
}

当我运行此程序时，在第一个节点和第二个节点的情况下崩溃。对于其他节点，则给出无限循环输出。 （例如：-2-> 4 2-> 4 2-> 4 ....等等）。

When I run this program, in case of 1st and 2nd node, it crashes. while in case of any other nodes, it gives infinite loop output. (eg:- 2->4 2->4 2->4....so on)`.

我知道关于节点交换还有更多问题，但没有找到与我的问题类似的任何东西。请帮帮我.. !!

I know there are some more questions about node swappings, but I didn't find any one similar to my problem. Please help me out..!!

预先感谢。

推荐答案

如果ptr1 == head（ptr1-> prev == NULL）或ptr2 == head（ptr2-> prev == NULL），则代码将失败，因为它最终会尝试使用head-> next，而不会不存在。如果ptr1-> next == NULL或ptr2-> next == NULL，也需要检查列表的末尾，可以使用本地尾指针进行处理。使用指向节点的指针可以简化代码。例如，指向ptr1的下一个指针的指针可以是& ptr1-> prev-> next或& head。指向ptr2的上一个指针的指针可以是& ptr2-> next-> prev或& tail（并设置tail = ptr2）。

The code will fail if ptr1 == head (ptr1->prev == NULL) or ptr2 == head (ptr2->prev == NULL), because it ends up trying to use head->next, which doesn't exist. There also needs to be a check for the end of a list, if ptr1->next == NULL or ptr2->next == NULL, which can be handled using a local tail pointer. Using pointers to pointer to node can simplify the code. For example the pointer to next pointer to ptr1 could be &ptr1->prev->next or &head. The pointer to prev pointer to ptr2 could be &ptr2->next->prev or &tail (and set tail = ptr2).

使用指向的指针节点解决了交换相邻节点的问题。

Using pointers to pointer to node fixes the issue with swapping adjacent nodes. Also temp can be a pointer to node.

示例代码使用指向节点的指针（而不是计数）进行交换：

Example code using pointers to nodes (instead of counts) to swap:

typedef struct node NODE;
/* ... */
NODE * SwapNodes(NODE *head, NODE *ptr1, NODE *ptr2)
{
NODE **p1pn;            /* & ptr1->prev->next */
NODE **p1np;            /* & ptr1->next->prev */
NODE **p2pn;            /* & b->prev->next */
NODE **p2np;            /* & b->next->prev */
NODE *tail;             /* only used when x->next == NULL */
NODE *temp;             /* temp */
    if(head == NULL || ptr1 == NULL || ptr2 == NULL || ptr1 == ptr2)
        return head;
    if(head == ptr1)
        p1pn = &head;
    else
        p1pn = &ptr1->prev->next;
    if(head == ptr2)
        p2pn = &head;
    else
        p2pn = &ptr2->prev->next;
    if(ptr1->next == NULL){
        p1np = &tail;
        tail = ptr1;
    } else
        p1np = &ptr1->next->prev;
    if(ptr2->next == NULL){
        p2np = &tail;
        tail = ptr2;
    }else
        p2np = &ptr2->next->prev;
    *p1pn = ptr2;
    *p1np = ptr2;
    *p2pn = ptr1;
    *p2np = ptr1;
    temp = ptr1->prev;
    ptr1->prev = ptr2->prev;
    ptr2->prev = temp;
    temp = ptr1->next;
    ptr1->next = ptr2->next;
    ptr2->next = temp;
    return head;
}

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