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

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 along with the first pointer (ptr1) until 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 the ptr2



c code for the above algorithm:

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);
}