在链表中交换2个节点C ++

Question

我试图交换一个链表中的2个相邻节点（在本例中为75和9），每次运行代码时，说一个链表75-> 9-> 767-> 2 ...，我得到的整个列表只是转到75-> 9-> 75-> 9-> 75-> 9等。我必须更新指针

Answer 1

Your problem is with the lines

nextPtr->next = tempPtr; nextPtr = tempPtr;

I'm not sure, but i think you only mean to type

nextPtr = tempPtr;

instead.

Answer 2

Hint: If you want to swap nodes H (what you're calling hdList) and N (nextPtr) in a linked list you have to make whatever points to H now point to N. You're not keeping track of whatever points to H at all.

That is, suppose you know that part of your list is

...   P -> H -> N -> Q ...

and you want to swap H and N. The state of the list after the swap should be

...  P -> N -> H -> Q ... 

right? But you can't do this because you don't know what used to point to H (ie P) so you can make it point to N.

I think you're going to have to go back to the drawing board on this one.

Answer 3

If you want to swap 2 nodes in linked list, why are you trying to swap the pointers and all? Just swap the data in it with the simple swap logic. let the pointers be as it is. If this is not what you want, tell me in detail what exactly you want it to be.

Answer 4

I guess you are looking for pairwise swap, use this for the same. Let me know if it went fine. for 75->9->767->2 it produces 9->75->2->762... If you want something different but similar, you can use it and make changes accordingly. void swap(struct node **head) { if (*head == NULL || (*head)->next == NULL) return;

    struct node *prev = *head;
    struct node *cur = (*head)->next;

    *head = cur;  

    while (true)
    {
        struct node *next = cur->next;
        cur->next = prev; 

        if (next == NULL || next->next == NULL)
        {
            prev->next = next;
            break;
        }

        prev->next = next->next;

        prev = next;
        cur = prev->next;
    }
}