@nishant, you wrote "until both the distance becomes equal".Which distances
? Could you please elaborate ?

On Thu, Jul 5, 2012 at 12:52 PM, Ashish Goel <[email protected]> wrote:

> struct node* intersection( struct node *pL1, struct node* pL2)
> {
>    if ((!pL1) || (!pl2)) return NULL;
>    struct node * pL3 = NULL;
>    struct node* pL3Tail = NULL;
>    while(pL1)&&(pL2) {
>     if (pL1->data< pL2->data) pL1=pL1->next;
>     else if  (pL1->data > pL2->data) pL2=pL2->next;
>     else {
>        struct node *pNew = (struct node*)malloc(sizeof(struct node));
>        if !pNew return NULL; //scary
>        pNew->data = pL1->data; pNew->next = NULL;
>        if ( !pL3) pL3= pNew;
>        else pL3Tail->next = pNew;
>        pL3Tail = pNew;
>    }
>    return pL3;
> }
>
>
>
>
> }
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Wed, Jul 4, 2012 at 10:41 PM, Abhi <[email protected]> wrote:
>
>> Any efficient algorithm to find intersection of two linked lists.Example:
>> Linked List 1)  1 -> 2 -> 3 -> 4 -> 5 -> 6
>> Linked List 2)  3 -> 4 -> 5
>>
>> Intersection 4 -> 5 -> 6
>>
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-- 
Abhishek Sharma
Under-Graduate Student,
PEC University of Technology

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