Question

Describe in detail how to swap two nodes x and y (and not just their contents) in a singly linked list L given references only to x and y. Repeat this exercise for the case when L is a doubly-linked list. Which algorithm takes more time. C++ please

Answer 1

//Swapping in singly Linked List

void swappingNodes(Node **head, Node* x, Node* y)
{
//if both nodes are same
if (x == y) return;

// Search for y (keep track of previous of node y and Current node y )
Node *prevY = NULL, *currentY = *head;
while (currentY && currentY!= y)
{
prevY = currentY;
currentY = currentY->next;
}

  
// Search for x (keep track of previous of node x and Current node x )
Node *prevX = NULL, *currentX = *head;
while (currentX && currentX != x)
{
prevX = currentX;
currentX = currentX->next;
}
  

// If either x or y is not present in the list
if (currentX == NULL || currentY == NULL)
return;
  
// If x is not head of linked list
if (prevX != NULL) prevX->next = currentY;
else // Else make y as new head
*head = currentY;
  
// If y is not head of linked list
if (prevY != NULL)
prevY->next = currentX;
else // Else make x as new head
*head = currentX;
  
// Swap next pointers
Node *temp = currentY->next;
currentY->next = currentX->next;
currentX->next = temp;
}

//swapping nodes in doubly linked list

void swapingNodes(struct Node* x, struct Node* y){
   if ( x->prev != NULL){
   x->prev->next = y;
   }
   else {
   head = y;
   }
  
if ( y->next !=NULL){
y->next->prev = x;
}
  
x->next = y->next;
y->prev = x->prev;
y->next = x;
x->prev = y;
}

swapping node in singly linked list takes more time.