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Microsoft Interview Question for Software Engineer/Developer (Fresher) about Linked Lists

(10 posts)

1. 

   akash
   guest
   Posted 3 months ago #
   
   

   Reverse a Linked List in size of 2

   Say LL is

   1->2->3->4->5->6->7->8

   Then u need to return

   7->8->5->6->3->4->1->2

2. 

   kartik
   Moderator
   Posted 3 months ago #
   
   

   1) Reverse the Linked List (See http://www.geeksforgeeks.org/archives/860)
   2) Swap the elements in pairs (See http://www.geeksforgeeks.org/archives/7049)

3. 

   camster
   Member
   Posted 3 months ago #
   
   

   @Kartik, Here is a recursive function that reverses Linked List in size 2 increments in -- 1 step only instead of two steps. Thank you. Camster

   node* reversesize2(node* LL){
   node* retval;
   if (LL && LL->next && LL->next->next == NULL){
   return LL;
   }
   if (LL && LL->next){
   retval = reversesize2(LL->next->next);
   swap(&LL,&(LL->next));
   node* x = retval;
   while (x->next){
   x = x->next;
   }
   if (x){
   x->next = LL->next;
   x->next->next = LL;
   if (x->next->next){
   x->next->next->next = NULL;
   }
   }
   return retval;
   }
   }

   void swap(node** node1, node** node2){
   int tmp;
   tmp = (*node1)->value;
   (*node1)->value = (*node2)->value;
   (*node2)->value = tmp;
   }

4. 

   Dedicated Programmer
   Member
   Posted 2 months ago #
   
   

   @camster ,can u pls explain your logic.

5. 

   elenashutova
   Member
   Posted 2 months ago #
   
   

   LList* reverse2(LList* head)
   {
   	if(!head)
   		return NULL;
   
   	LList* res = NULL;
   	LList* cur = head;
   	LList* nextnext = (cur->next ? cur->next->next : NULL);
   
   	while(cur)
   	{
   		nextnext = (cur->next ? cur->next->next : NULL);
   		if(cur->next)
   		{
   			cur->next->next = res;
   		}
   		else
   		{
   			cur->next = res;
   		}
   		res	= cur;
   		cur = nextnext;
   	}
   	return res;
   }

6. 

   camster
   Member
   Posted 1 month ago #
   
   

   dedicated_programmer, reversesize2 uses the concept of induction. reversesize2(LL->next->next) finds the reverse linked list of size 2 of LL->next->next and then connects the return value of
   reversesize2(LL).specifically returnvalue->next->next, to the return value of reversesize2(LL->next->next). It is not very complicated. Thank you Camster,

7. 

   Dedicated Programmer
   Member
   Posted 1 month ago #
   
   

   Initially,root and newroot are root nodes
   

   struct node* reverse(struct node *root,struct node **newroot)
   {
   	struct node *temp,*start;
   	if(!root)
   		return NULL;
   	if(root->next && root->next->next)
   	{
   		temp=root;
   		start=reverse(root->next->next,newroot);
   
   		start->next->next=temp;
   
   		temp->next->next=NULL;
   	}
   	else
   		*newroot=root;
   	return root;
   }

8. 

   camster
   Member
   Posted 1 month ago #
   
   

   Dedicated_Programmer. I just tested your reverse function on the test case below and your reverse function returns the wrong value. Plz fix it. Thank you.

   Say LL is

   1->2->3->4->5->6->7->8

   Then u need to return

   7->8->5->6->3->4->1->2

9. 

   Dedicated Programmer
   Member
   Posted 1 month ago #
   
   

   camster, I have again checked the code, but it is returning the output 7->8->5->6->3->4->1->2 .
   As i have mentioned newroot contains the address of the root node.
   Pls check it. I am waiting to hear from you.
   Thank you.
   Dedicated Programmer

10. 

    camster
    Member
    Posted 1 month ago #
    
    

    Dedicated Programmer, I tested your code. You proposed solution modifies the pointer to pointer new root. IMHO, Ny solution is better than yours because I return the reversed size-2 list wheras your return value is returns 7->8->NULL. Plz correct ur solution or elese I can do it for you ig you pay me $50/hour. Thank you,

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