both the distance means distance between the two linked list some thing
like this :
struct node *temp1,*temp2;
temp1 = head1;
temp2 = head2;
int l1 = get_dist(head1);
int l2 = get_dist(head2);
int ds = abs(l1-l2);
if( l1 > l2 )
{
for(int i=0;i<ds;i++)
{
if(temp1->next)temp1 = temp1->next;
}
// here distance of both the list is equal .
// now you may traverse like this
while(temp1 && temp2)
{
if(temp1 == temp2 )
// intersection point
else
temp1 = temp1->next;
temp2 = temp2->next;
// move both one one step a time will come both will meet the
intersection point
// i hope the point is clear if not let me knw i will send u
the whole code .
}
}
else
{
// similary for (l1<l2)
}
On Thu, Jul 5, 2012 at 4:32 PM, Abhishek Sharma <abhi120...@gmail.com>wrote:
> @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 <ashg...@gmail.com> 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 <abhi120...@gmail.com> 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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