The following code is generating a runtime error in one type of testcase.

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In this question, I need to remove the duplicate elements of the linked list. But it is generating error in some of the test cases(to be specific in those cases where the duplicate element occurs at the end of the linked list). Can someone help me figure out the error?

What I have tried:

SinglyLinkedListNode* removeDuplicates(SinglyLinkedListNode* llist)
{
    if(llist==NULL)
    return NULL;
    
    SinglyLinkedListNode* curr=llist;
    while(curr!=NULL && curr->next!=NULL)
    {
        SinglyLinkedListNode* temp=curr;
        while(temp->next!=NULL)
        {
            if(temp!=NULL && temp->next->data==curr->data)
            {
                SinglyLinkedListNode* nodetodelete=temp->next;
                temp->next=temp->next->next;
                delete(nodetodelete);
            }
            temp=temp->next;
        }
        curr=curr->next;
    }
    return llist;
}

Posted 21-Aug-22 2:13am

Member 15079126

Updated 22-Aug-22 16:42pm

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Rick York 21-Aug-22 22:17pm

In my opinion, the easiest way is to not let them be added in the first place.

2 solutions

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OriginalGriff · Answer 1 · 2022-08-21T02:25:00

At the end of the list, temp->next will be null to indicate the end of the list.
So this code:

C++

if(temp!=NULL && temp->next->data==curr->data)

will always fail for the last item as the next item doesn't exist.

Frankly, I'm not at all sure what temp->next->data==curr->data is supposed to be checking for, so I can't suggest a fix. I'd go back to the original instructions and check that part fairly carefully.

merano99 · Answer 2 · 2022-08-22T16:42:00

Solution 2

Shouldn't the reference to the next element be better in the else branch?

C

if(temp!=NULL && temp->next->data==curr->data)
{
  SinglyLinkedListNode* nodetodelete=temp->next;
  temp->next=temp->next->next;
  delete(nodetodelete);
}
else
  temp=temp->next;

Posted 22-Aug-22 16:42pm

merano99

Updated 22-Aug-22 16:43pm

v2