1-2-3-4-5-6-7-8-9-10-11
fast and slow meet at 11
m=6;
k=4
...i cant get last two lineswhen k= sometimes around the circle -
m..
then after that taking fast at begining and slow within circle ..i cant get
this...@ dave plz explain with this example...will b of gr8 help..thnx in
@Rahul:
As Dev said:
Now start the fast pointer at the head and take m single steps with both
pointers. The fast pointer is at the beginning of the cycle, and the slow
pointer has traversed the cycle (2*t - u) times and is back at the
beginning of the cycle.
k= (2*t-u)p -m
When fast pointer is
@moheed i got it.
@Dave...Thnx a lot..i got it all now...
On Sat, Mar 10, 2012 at 5:56 PM, Moheed Moheed Ahmad mohe...@gmail.comwrote:
@Rahul:
As Dev said:
Now start the fast pointer at the head and take m single steps with both
pointers. The fast pointer is at the beginning of the cycle,
i have 2 pointers fast and slow.now if tehy meet there is a loop...
now keep one ptr at meeting point and take other one to the begining of
listmove both at speed of one..they will meet at start of loophow
this happens???why they meet at start..plz tell logic behind this???thnx in
No They will not meet at the start in a case containing 5 nods and having
loop at the third node. once check this
On Fri, Mar 9, 2012 at 3:48 PM, rahul sharma rahul23111...@gmail.comwrote:
i have 2 pointers fast and slow.now if tehy meet there is a loop...
now keep one ptr at meeting
@ rahul sharma:
the linked list is a combination of a list a-b-...-p-q and a cycle
q-r-...-z-q. (z != p).
noting that the start of cycle q is the only node with 2 predecessor: p
and z.
if 2 pointers meet at some node x, different from q, in last step they
must have met at x', the predecessor
@terencei cant get..can u eleboratethnx for the sol..but plz
elaborate...
On Fri, Mar 9, 2012 at 5:59 PM, Terence technic@gmail.com wrote:
@ rahul sharma:
the linked list is a combination of a list a-b-...-p-q and a cycle
q-r-...-z-q. (z != p).
noting that the start of cycle q
suppose linked list is
a-b-c-d-e
and suppose loop starts from 'c'
according to u let one pointer be at 'c' say *q and another be at 'a' say
*p. Now if we move both at the speed of one then
After first pass
p will be at b
q will be at d
After second pass
p will be at c
q will be at e
@sanjiv...how can u take q at c.they will meet at some position ...
suppose u have
1-2-3-4-5-6-7-4
now initialy
slow and fast both at 1
start incrementing slow by 1 and fast by 2
slow @2 ...fast @3
slow @3fast @5
s...@4...fast@7
slow @5...fast@5
both are equalmeans dere is
I think u r making mistake i.e. after 7,5 will com not 4 because loop
starts from 5.check it again
On 3/9/12, rahul sharma rahul23111...@gmail.com wrote:
@sanjiv...how can u take q at c.they will meet at some position ...
suppose u have
1-2-3-4-5-6-7-4
now initialy
slow and fast
itz ryt...loop starts from 4 n not from 5...
On Fri, Mar 9, 2012 at 8:58 PM, sanjiv yadav sanjiv2009...@gmail.comwrote:
I think u r making mistake i.e. after 7,5 will com not 4 because loop
starts from 5.check it again
On 3/9/12, rahul sharma rahul23111...@gmail.com wrote:
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