swapping adjacent nodes in a Linked list

Question

I have problems swapping adjacent nodes in a linkedlist.

for ex: input : 1->2->3->4->5->null output: 2->1->4->3->5->null

bool swapAdjacent(node** head)
{

//1->2->3->4->null

//2->1->4->3->null
if(head==NULL)
return 0;
node* current  = *head;
*head = (*head)->next ;
node* prev = NULL;
cout<<"head val "<<(*head)->data <<endl;
node* temp;
while( current!=NULL&&current->next!=NULL)
{
   temp = current->next ;  //1s pointer points to 2
   current->next = temp->next ;    // 1s pointer point to 3
   temp ->next = current;   //2s pointer shud point to 1
   prev = current;
   current = current->next ;
   //cout<<"data " <<current->data <<endl;

   if(current!=NULL)
   prev->next = current->next ;



}

return 1;
}

My code is not working whenever there are odd no of nodes. How to fix this ?

Answer 1

Why so complicated?

int swapAdjacent(node** head) {
  if (!*head || !(*head)->next)
    return 0;
  node* sw = (*head)->next;
  (*head)->next = sw->next;
  sw->next = *head;
  *head = sw;
  swapAdjacent(&(sw->next->next));
  return 1;
}

Edit: changed return value.

Answer 2

I don't know what is wrong with your code, but I would approach it by showing the state the of list at each pass through the loop:

void printList(node* head)
{
    int cntr = 1;

    while (head != NULL)
    {
        printf("(Node: %d. Name: %s) --> ", cntr, head->name);
        head = head->next;
    }
}

For this, make sure each node has a name, set to "1", "2", etc. Then printout the state of the list on each pass, perhaps even each step! And you should see what's going wrong.