Circular linked list

#1

sir want a hint plz to solve this problem.

#2

@sameeksha,

Say If INPUT IS 1 2 3 4 5 2 3 -1

Form a linked list 1->2->3->4>5 then you see that 2 has already been visited , so connect the next of 5 to 2 (which has already been created). This way you have created the circular linked list . After that just detect the cycle, remove the cycle, print the new linked list.

Maintain a map or a visited array…whenever you see that a node is already visited, connect next of previous node to that node

On how to remove the circular loop:

We will use Floyd Cycle Detection algorithm to detect and remove the loop. In the Floyd’s algo, the slow and fast pointers meet at a loop node. We can use this loop node to remove cycle. There are following two different ways of removing loop when Floyd’s algorithm is used for Loop detection.

Method 1 (Check one by one):

  1. We know that Floyd’s Cycle detection algorithm terminates when fast and slow pointers meet at a common point. We also know that this common point is one of the loop nodes (2 or 3 or 4 or 5 in the above diagram).

  2. We store the address of this in a pointer variable say ptr2.

  3. Then we start from the head of the Linked List and check for nodes one by one if they are reachable from ptr2.

  4. When we find a node that is reachable, we know that this node is the starting node of the loop in Linked List and we can get pointer to the previous of this node.

Method 2 (Better Solution) :
This method is also dependent on Floyd’s Cycle detection algorithm.

  1. Detect Loop using Floyd’s Cycle detection algo and get the pointer to a loop node.
  2. Count the number of nodes in loop. Let the count be k.
  3. Fix one pointer to the head and another to kth node from head.
  4. Move both pointers at the same pace, they will meet at loop starting node.
  5. Get pointer to the last node of loop and make next of it as NULL.