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

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);
}