sorry, pls ignore above code ..a bit modifie code
I'm passing length = 0
printPath (int length)
{
if(lenght == N) return;
else
{
if(lenght+1 <= N)
{ Printf (" -->one step -->") call printPath (length+1) }
else if(lenght+2 <= N)
{ Printf (" -->two steps -->") call printPath (length+2) }
else
return;
}
On Sun, Aug 28, 2011 at 6:54 PM, Kamakshii Aggarwal
<[email protected]>wrote:
> what is the initial value of length that u r passing??
>
> On Sun, Aug 28, 2011 at 6:47 PM, prashant thorat <[email protected]
> > wrote:
>
>> Hi, you can do this by recursion ,
>>
>> printPath (int length)
>> {
>> if(lenght == N) return;
>> else
>> {
>> if(lenght+1 >= N) { Printf (" -->one step -->") call printPath
>> (length+1) }
>> else if(lenght+2 >= N) { Printf (" -->two steps -->") call printPath
>> (length+2) }
>> else
>> return;
>>
>> }
>> }
>> On Sun, Aug 28, 2011 at 5:42 PM, himanshu kansal <
>> [email protected]> wrote:
>>
>>> you have a path of N steps....
>>> at every step, you can take onl;y 1 step or 2 steps.....
>>>
>>> how to print all the possible paths that you can take......
>>>
>>> PS: please dont give the exact code.....your approaches will be
>>> appreciated....:)
>>>
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>>>
>>
>>
>> --
>> Yours affectionately,
>> Prashant Thorat
>> Computer Science and Engg. Dept,
>> NIT Durgapur.
>>
>> --
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>
>
>
> --
> Regards,
> Kamakshi
> [email protected]
>
> --
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> "Algorithm Geeks" group.
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>
--
Yours affectionately,
Prashant Thorat
Computer Science and Engg. Dept,
NIT Durgapur.
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