bits manipulation tricks is cool to solve these kind of questions.

On Sun, Dec 26, 2010 at 3:27 PM, Praveen Baskar <[email protected]>wrote:

> Your Second approach is cool :)
>
> On Wed, Dec 22, 2010 at 1:06 PM, juver++ <[email protected]> wrote:
>
>> Use bits manipulation tricks.
>> 1. There is a way to remove a group of consecutive 1's from the right: A =
>> n & (n + 1). Then check if A==0 then OK.
>> 2. Second approach: B=n+1, check if B & (B-1) (this checks if B is a power
>> of 2, so it contains only 1 set bit) is zero then OK.
>>
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>
>
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> B. Praveen
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