@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
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.comwrote:
On Sat, Apr 7, 2012 at 11:24 AM, Dave dave_and_da...@juno.com wrote:
Hello Dave,
Was trying this
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0; /*step1*/
n |= n-1;/*step 2*/
return !(n (n+1)); /*step 3*/
}
for 1110011
after step 1 it becomes 1110010100=111
after step2 it becomes 111|110=111(all ones)
got it...super...
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Thu, May 17, 2012 at 8:21 PM, Ashish Goel ashg...@gmail.com wrote:
Hello Dave,
Was trying this
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return
Hii Can anyone explain to me what this code is doing??
On 17 May 2012 20:36, Ashish Goel ashg...@gmail.com wrote:
got it...super...
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Thu, May 17, 2012 at 8:21 PM, Ashish Goel
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
Hello Ashish,
I would do something like here -
http://www.evaluzio.net/library/verify-1-0-bits-in-num.htm
Basically you just count the number of 01 and 10 binary sequences using (n
3 == 1) or (n 3 == 2)
Regards,
Ilya
On Tuesday, 3 April 2012 17:00:36 UTC-7, ashgoel wrote:
verify that
@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
