Comments on: Interview Questions: Loops in Linked Lists

By: 800 calorie diet

800 calorie diet — Tue, 13 Jul 2010 01:17:40 +0000

Golden. Great, useful info.

By: Air Climber

Air Climber — Sat, 03 Jul 2010 21:36:13 +0000

No this has to be a web programming language

By: Magic Jack

Magic Jack — Fri, 02 Jul 2010 20:07:53 +0000

Is this for C++?

By: vidarislying

vidarislying — Thu, 01 Jul 2010 18:22:54 +0000

Sorry to say vidar you're blowing hot air out some orifice! the tortoise and hare solution you claim to have “reasoned” out in an interview was invented by Bob Floyd who won the Turing award which means he was a computer science genius – I really doubt even he figured it out in 15 minutes which is by far the maximum time interviewers would allow you to solve a problem. People like you who pretend to be smart are the worst interviewers because you lie to yourself about your abilities and make job candidates suffer for it.

By: gournew

gournew — Thu, 08 Apr 2010 11:43:31 +0000

How to break the loop:
step1:
Detect the loop using slow(p1) and first pointer(p2)
step2:-
keep one poionter (p1)there and reset pointer (p2=head)to head.
step 3:
go still both are same now make p2->next=NULL.
complete code::-
http://goursaha.freeoda.com/DataStructure/LoopD...

By: axvpast

axvpast — Tue, 23 Feb 2010 02:45:58 +0000

For my opinion the best solution is just extend Node by additional attribute (visited).
After that we have linear algorithm complexity without performance and resources problems. In all other cases we need more memory or more CPU time.

By: jc

jc — Fri, 06 Jun 2008 16:09:01 +0000

In the ‘easy solution’ you also first need to check for a list that has 2 nodes that are in loop.

By: Jesse

Jesse — Wed, 14 May 2008 17:14:49 +0000

Shaneal,

You’re right.

By: Shaneal Manek

Shaneal Manek — Wed, 14 May 2008 12:51:53 +0000

I believe you have an error in your analysis of the ‘easy solution.’

It runs in O(n^2) time, not O(n). I think since you are using an array to store the nodes already seen,for an n item list it will take take a total of
n + (n-1) + (n-2) + … + 1 operations,
to determine if a given node has already been seen.

This is an arithmetic progression sums to n*(n+1)/2 = 1/2*(n^2+n) which is O(n^2) complexity.

I believe you can get an O(n) time complexity if you use a hash table to store the nodes already seen instead of an array.

By: Jesse

Jesse — Tue, 29 Apr 2008 05:35:35 +0000

Jim,

1) I never said it ran in constant time. I said it took constant memory, which is true.

2) has_loop? expects a Node object to be passed in, so it should never be nil. In a real-world situation I’d of course have exception handling, but the examples were meant to be illustrative rather than complete.

Cheers,
Jesse