@Ashgoel: My solution worked when the pattern of ones followed by zeros
followed by ones is right-justified in a larger integer, so that there are
zeros to the left of the leftmost one. E.g., my algorithm will say that
00110011 in an 8-bit integer is valid.
Dave
On Friday, May 18, 2012 1:23:28 AM UTC-5, ashgoel wrote:
> i changed the last step to return (n == ~0);
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Fri, May 18, 2012 at 11:27 AM, Vishal Thanki <vishaltha...@gmail.com>wrote:
>
>> On Sat, Apr 7, 2012 at 11:24 AM, Dave <dave_and_da...@juno.com> wrote:
>> > @Ashgoel: Try this:
>> >
>> > bool OnesZerosOnes(unsigned int n)
>> > {
>> > if( !(n & 1) || !(n &= n+1) ) return 0;
>> > n |= n-1;
>> > return !(n & (n+1));
>> > }
>> >
>> > Here is how it works:
>> >
>> > !(n & 1) is true if the number has trailing zeros.
>> >
>> > If the number has trailing ones, n &= n+1 replaces the trailing ones
>> with
>> > zeros.
>> >
>> > !(n &= n+1) is true if there are only trailing ones, i.e., the original
>> > number was zeros followed by ones.
>> >
>> > n |= n-1 replaces trailing zeros with ones. Thus, if the original
>> number is
>> > ones followed by zeros followed by ones, the zeros have been changed to
>> > ones.
>> >
>> > (n & (n+1)) replaces the trailing ones with zeros. If the number is now
>> > zero, the number is valid, otherwise the number is invalid.
>> >
>> > Dave
>> >
>> >
>>
>> Superb, Dave!!
>>
>> > On Tuesday, April 3, 2012 7:00:36 PM UTC-5, ashgoel wrote:
>> >>
>> >> verify that the bits of a number are in format 1s followed by 0s
>> followed
>> >> by 1s like 1110001 is valid but 100100100 is not
>> >>
>> >> Best Regards
>> >> Ashish Goel
>> >> "Think positive and find fuel in failure"
>> >> +919985813081
>> >> +919966006652
>> >
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