在C中的双链表中交换节点

Question

I am trying to swap nodes in a double linked list in C.

My list if as it follows: M4,M3,M2,M1. I have a function that computes the dimension of the list and it reads it correctly as 4.

It all breaks down when I try to swap M3 with M2.(hoping to get after traversing it this: M4,M2,M3,M1.)

I use the following line in main: (prim is the the first node, M4)

swap_nodes1(prim->pNext, prim->pNext->pNext);

And the following code serves as my swapping function. (Note:i still need to cover the cases when switching the head with another node, or the tail with another node, yet I wanted first to swap some random middle positioned nodes.)

void swap_nodes1(pNODE object1, pNODE object2)
{
pNODE temp1,temp2;
temp1->pPrec=object1->pPrec;
temp1->pNext=object1->pNext;

object1->pPrec=object2->pPrec;
object1->pNext=object2->pNext;
object2->pPrec=temp1->pPrec;
object2->pNext=temp1->pNext;

free(temp1);
}

When I traverse the list for the 2nd time,i get: M4,M3,M1. Also the dim of the list is now 3 instead of 4,M2 magically disappearing.

Can anyone provide me an explanation on why I am getting this and what I am doing wrong?

Thank you in advance.

Answer 1

This is much easier if you draw it on paper and research it that way.

Assuming object1 and object2 are not NULL.

You need to think about the nodes that come before and after the two nodes you are swapping. The node just before your two swap nodes needs to point to object2 now instead of object1 (assuming that the node exists). And you need to make object2's preceding now point to the node just before your two swap nodes.

if ( object1->pPrec != NULL )
   object1->pPrec->pNext = object2;
object2->pPrec = object1->pPrec;

And you need to make object1's next points to the node just after your two swap nodes and that node needs to point back to object1 (if it exists).

object1->pNext = object2->pNext;
if ( object2->pNext != NULL ) 
    object2->pNext->pPrec = object1;

Then object1 needs to point back to object2 and object2 needs to point forward to object1.

object1->pPrec = object2;
object2->pNext = object1;

Answer 2

The problem statement doesn't specify what points to the first node of the list (M4), or if there is a pointer to the last node of the list (M1). It doesn't state if the list is circular, so I assume it's not circular.

When swapping nodes in a list, if the nodes to be swapped are adjacent (which is the example case, M3 and M2), then pointers need to be "rotated". If the nodes to be swapped are not adjacent, then pointers need to be swapped. The same logic can be used for both cases by first swapping what ever points to the two nodes, then swapping the two nodes pointers.

Using pointer to pointer to node will simplify the logic.