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# Data Structure/Algorithm Challenger II

We have a doubly-linked-list with number of nodes unknown.

Sometimes, while working with the list, it somehow gets corrupted. The corrupt node is pointing to some previous node instead of next node. So a circle is created in the list.

Test for the existence of circle (i.e., check if the list is corrupt).

Note: The list is doubly and not singly like Data Structure/Algorithm Challenger I, so a solution other then one specified in Data Structure/Algorithm Challenger I is expected.
sahana • Mar 16, 2008
some hint please. isnt the one by single linked list considered to be the most optimal?i jus hav a vague idea.may be we can have two pointers that go in opposite directions and if they dont meet means there is a cycle.i dont know where my second pointer ie which is to travel back wards has to point initially because i am not given the number of nodes.
sahana
isnt the one by single linked list considered to be the most optimal?
The solution given for singly linked list may be optimal for it, but a more optimized solution may exist for doubly linked list.

sahana
may be we can have two pointers that go in opposite directions and if they dont meet means there is a cycle.i dont know where my second pointer ie which is to travel back wards has to point initially because i am not given the number of nodes.
We can't have second pointer pointing to the last node of list because we have only head and if the list is corrupt we never reach the last node. Due to same reason we canit count the number of nodes in the list.

sahana
The basic property of doubly linked list is that every node has two pointers, one pointing to next node and other pointing to previous node in the list. Having an extra pointer in each node surely increase the space complexity, but that will help in giving more optimized solution to the problem in terms of space complexity.

No solution from such a long time 😔

Here is a sample code for the given problem statement. Solution is based on fact that in a doubly linked list, previous of next of a node is the same node. The solution has complexity of O(n).

```n[FONT=Verdana]ode *tmp1 = head;[/FONT]
[FONT=Verdana]int result = 0;        //Specify no circular path exist[/FONT]

[FONT=Verdana]while(tmp1 != NULL) {[/FONT]
[FONT=Verdana]  if(tmp1->next != null) {[/FONT]
[FONT=Verdana]      if(tmp1->next->prev != tmp1) {[/FONT]
[FONT=Verdana]          result = 1;        //specify existence of circular path[/FONT]
[FONT=Verdana]          break;[/FONT]
[FONT=Verdana]      }[/FONT]
[FONT=Verdana]      tmp1 = tmp1->next;[/FONT]
[FONT=Verdana]  } else {[/FONT]
[FONT=Verdana]      break;[/FONT]
[FONT=Verdana]  }[/FONT]
[FONT=Verdana]}[/FONT]

[FONT=Verdana]return result;[/FONT]
```