Implement a function that returns the 5th element from the
end in a singly linked list of integers in one pass.
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Answer / amol b
int return_fifth_from_end()
{
int a[5],curr_ct=0;
struct node *p;
p=head;
while(p->next!=NULL)
{
a[curr_ct%5]=p->val;
p=p->next;
curr_ct++;
}
if(curr_ct>=5)
return a[(curr_ct-5)%5];
else
return -1;
}
| Is This Answer Correct ? | 8 Yes | 0 No |
Answer / abhijit annaldas
Sorry, it was my mistake.. previous answer was not correct.
Here is the corrected one...
node* getNthFromLast(node* head, int n)
{
int c=0;
node *nth=head;
node *pt=head;
while(pt!=NULL)
{
pt=pt->next;
c++;
//if c=n then nth node is already set to head.
if(c>n)
nth=nth->next;
}
if(c<n) //LL contains less than n nodes
return (*node)0;
else
return *nth;
}
Use it as..
fifth_node = getNthFromLast(head, 5);
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / tarun dhiraj
Consider:
struct Node
{
int data;
struct Node *next;
}*start;
void FIFTHFRMLAST()
{
struct Node *ptr;
ptr=start;
printf("\n");
/*Traverse elements of linked list till the 5th element from
the end of linked list*/
while(ptr->next->next->next->next->next!=NULL)
{
ptr=ptr->next;
}
printf("->%d",ptr->data);
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / abhijit annaldas
node* getNthFromLast(node* head, int n)
{
int c;
node *nth;
while(pt!=NULL)
{
pt=pt->next;
c++;
if(c>=n)
*nth=pt;
}
if(c<n) //LL contains less than n nodes
return (*node)0;
else
return *nth;
}
Use it as..
fifth_node = getNthFromLast(head, 5);
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / manesh nambiar
int return_fifth_from_end()
{
int i,j;
struct node *p,*q;
*p=HEAD_NODE;
for(i=0;i<4;i++)//will make p point to the 5th element
{
p=p->next;
if(p==NULL)
{
printf("List has less than 5 elements");
}
}
q=HEAD_NODE;
while(p!=NULL)
{
p=p->next;
q=q->next;
}
return(q->Value);
}
| Is This Answer Correct ? | 8 Yes | 7 No |
Answer / lly
I think both #1 & #2 are wrong. Because the code travels
the list in two passes not one pass.
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / jishu
Answer 6 is correct and well thought but there is a tiny
mistake.
the while loop should check upto the last node which it
doesn't and hence prints the wrong value ie. value of the
node before the fifth node from the end.
while(p!=NULL) would give the correct node.
Also while returning the value, decrementing curr_ct by 5 is
not really necessary.
return a[curr_ct%5]; would give the same value in all cases.
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / vignesh1988i
let us assume that we have created a linked lists ..... the
no. of nodes is say 15.
count=15;
printf("enter the node do you wann to look out :\n");
scanf("%d",&i);
if(i>count || i>=0)
printf("no node exists like this \n");
else
{
i=count-i;
view_node(i);
printf("%d",HEAD1->data);
}
getch();
}
void view_node(int i)
{
HEAD1=HEAD;
for(int k=1;k<i;k++)
HEAD1=->HEAD1->ptr;
}
| Is This Answer Correct ? | 2 Yes | 2 No |
Answer / gbohrn
int return_fifth_from_end()
{
int i,j;
struct node *p,*q;
*p=HEAD_NODE;
for(i=0;i<4;i++)//will make p point to the 5th element
{
p=p->next;
if(p==NULL)
{
printf("List has less than 5 elements");
}
}
q=HEAD_NODE;
while(p->next!=NULL)
{
p=p->next;
q=q->next;
}
return(q->Value);
}
| Is This Answer Correct ? | 5 Yes | 6 No |
Answer / a.eklare7
int fifth_element_last(node *head)
{
node *p;
int len=0;
p=head;
while(p->next!=NULL)
{
p=p->next;
len++;
}
p=head;
for(int i=1;i<=len-4;i++)
p=p->next;
return(p->data);
}
| Is This Answer Correct ? | 0 Yes | 1 No |
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