Friday, May 25, 2007

Write a C program to return the nth node from the end of a linked list.

Here is a solution which is often called as the solution that uses frames.

Suppose one needs to get to the 6th node from the end in this LL. First, just keep on incrementing the first pointer (ptr1) till the number of increments cross n (which is 6 in this case)


STEP 1 : 1(ptr1,ptr2) -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10

STEP 2 : 1(ptr2) -> 2 -> 3 -> 4 -> 5 -> 6(ptr1) -> 7 -> 8 -> 9 -> 10



Now, start the second pointer (ptr2) and keep on incrementing it till the first pointer (ptr1) reaches the end of the LL.


STEP 3 : 1 -> 2 -> 3 -> 4(ptr2) -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 (ptr1)


So here you have!, the 6th node from the end pointed to by ptr2!


Here is some C code..


struct node
{
int data;
struct node *next;
}mynode;


mynode * nthNode(mynode *head, int n /*pass 0 for last node*/)
{
mynode *ptr1,*ptr2;
int count;

if(!head)
{
return(NULL);
}

ptr1 = head;
ptr2 = head;
count = 0;

while(count < n)
{
count++;
if((ptr1=ptr1->next)==NULL)
{
//Length of the linked list less than n. Error.
return(NULL);
}
}

while((ptr1=ptr1->next)!=NULL)
{
ptr2=ptr2->next;
}

return(ptr2);
}

32 comments:

Ankit said...

good solution man

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Anonymous said...

one could reverse the list and start...

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Chandra said...

Try it!!!

int nthnode(struct node *p, int n)
{
if( p == NULL)
return 0;
else
{
int x = 1+nthnode((p->next),n);
if( x == n)
cout<<"Nth Element: "<i<<endl;
return x;
}
}

Thanks,
Chandra

Chandra said...

int nthnode(struct node *p, int n)
{
if( p == NULL)
return 0;
else
{
int x = 1+nthnode((p->next),n);
if( x == n)
cout<<"Nth Ele: "<i<<endl;
return x;
}
}

Chandra said...

int nthnode(struct node *p, int n)
{
if( p == NULL)
return 0;
else
{
int x = 1+nthnode((p->next),n);
if( x == n)
{
cout<<"Nth Ele: "<<
p->i<<endl;
}
return x;
}
}

Palmer Eldritch said...

Can be optimized easily, instead of using two loops - merge them into one by adding one condition which checks counter.

As alternative solution - one can first count the length of LL, and then search for (count - n)th element.
second alternative is reversing LL and get nth element from the start of LL.

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Kinshuk Chandra said...

I also like the 2 pointer approach. But there is a recursive solution to this as well as written here - http://k2code.blogspot.in/2010/04/return-nth-node-from-end-of-linked-list.html:

node* findNthNode (node* head, int find, int& found){
if(!head) {
found = 1;
return 0;
}
node* retval = findNthNode(head->next, find, found);
if(found==find)
retval = head;
found = found + 1;
return retval;
}