MESSAGE
| DATE | 2015-03-01 |
| FROM | Ruben
|
| SUBJECT | Subject: [LIU Comp Sci] Fwd: Re: Linked List Test Question
|
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Message-ID: <54F352D2.60702-at-my.liu.edu>
Date: Sun, 01 Mar 2015 12:56:34 -0500
From: Ruben
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To: learn-at-nylxs.com
Subject: [LIU Comp Sci] Fwd: Re: Linked List Test Question
References: <781503C8-CEB4-4856-A980-10A1C75C54E4-at-liu.edu>
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-------- Forwarded Message --------
Subject: Re: Linked List Test Question
Date: Sun, 1 Mar 2015 12:18:10 +0000
From: Jose A. Rodriguez
To: Ruben
The question does not refer to any algorithm in particular. It just refers to the situation depicted. Pointer variable q is currently pointing to node containing 20. Pointer variable r is currently pointing to node containing 30. Pointer variable n is pointing to the new node to be inserted. So
q -> next = n; n->next = r;
is perfectly fine. So would be
n -> next = r; q-> next = n;
if that had been a choice.
How I got q and r to point to 20 and 30 is not relevant to the question.
> On Feb 28, 2015, at 11:54 PM, "Ruben" wrote:
>
>
> I'm looking at this question and I don't think it is right. In
> theory it is correct but in reality with the algorithm, you can't
> detach the upper part of the linked list and just assign it
> because it then ends up as lost pointer and leaked memory.
>
>
>
> That is why you always assign the value of q->next to your new
> node FIRST. Now you have a handle on the upper part of the list
> and you can now assign q->next with your new node address and your
> finished.
>
>
>
>
> Question 5
>
> 5 out of 10 points
>
>
>
> Partial Credit
>
> Consider the linked list in the handout. Which of the following
> correctly inserts node containing 25 between the nodes containing 20 and 30?
>
>
>
> Correctd.
>
> n->next = q->next;
>
> q->next = n;
>
> Answers:
> Correcta.
>
> q->next = n;
>
> n->next = r;
>
>
>
> b.
>
> p->next = n;
>
> n->next = r;
>
>
>
> c.
>
> q->next = n;
>
> n->next = q->next;
>
>
>
> Correctd.
>
> n->next = q->next;
>
> q->next = n;
>
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