ya that's a better and simplified approach....but i was trying to clear
rahul doubt's bcoz he was not moving the second pointer...i gave
explanation for that
--


Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad
<http://gplus.to/amolsharma99>
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On Sun, Nov 20, 2011 at 2:38 PM, kartik sachan <[email protected]>wrote:

> @amol..we can simply break the loop where we find the slow and fast
> pointer meeting,then its become a a problem of two link list having common
> ending point and different staring point
> and for detecting the extact point where loop began is two traverse both
> the list and find its length lets say 'a' and 'b' now put 1st pointer at
> start position of 1st link list and other pointer at distance of |a-b| from
> second list ...now where both of them meet is the point where loop began.
>
> correct me if i am worng ...
>
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