Reputation: 756
In the interview question, "Implement an algorithm which detects for presence of the loop.". For example, the linked list has a loop, like:
0--->1---->2---->3---->4---->5---->6
▲ |
| ▼
11<—-22<—-12<—-9<—-8
Using Floyd's cycle detection, this problem can be solved by using a fast & slow pointer. So should I try comparing
a. Link's node values, i.e.
if (fast.data == slow.data)
break;
where fast and slow are of type Link
class Link
{
int IData {get; set;}
Link Next {get; set;}
}
OR
b. Are they pointing to same reference, i.e. if (fast == slow)
Thanks.
Upvotes: 0
Views: 6011
Reputation: 460
Hope this helps... It might be naive but it works...
using System;
namespace CSharpTestTemplates
{
class LinkedList
{
Node Head;
public class Node
{
public int value;
public Node NextNode;
public Node(int value)
{
this.value = value;
}
}
public LinkedList(Node head)
{
this.Head = head;
}
public Boolean hasLoop()
{
Node tempNode = Head;
Node tempNode1 = Head.NextNode;
while(tempNode!=null && tempNode1!=null){
if(tempNode.Equals(tempNode1)){
return true;
}
if ((tempNode1.NextNode != null) && (tempNode.NextNode != null))
{
tempNode1 = tempNode1.NextNode.NextNode;
tempNode = tempNode.NextNode;
}
else
{
return false;
}
}
return false;
}
public static void Main()
{
Node head = new Node(1);
LinkedList ll = new LinkedList(head);
Node node2 = new Node(2);
Node node3 = new Node(3);
Node node4 = new Node(4);
Node node5 = new Node(5);
Node node6 = new Node(6);
head.NextNode = node2;
node2.NextNode = node3;
node3.NextNode = node4;
node4.NextNode = node5;
node5.NextNode = node6;
node6.NextNode = null;
Console.WriteLine(ll.hasLoop());
Console.Read();
}
}
}
Upvotes: 0
Reputation: 1503479
You should only be comparing the nodes themselves. After all, it's reasonable to have a linked list with repeated data in, without it actually having a cycle.
I would call them nodes rather than links too. A link is simply the reference from one node to the next or previous one - in particular, there's no data associated with a link, only with a node.
Upvotes: 8