On Mar 21, 7:40 am, "hijkl" <[EMAIL PROTECTED]> wrote:
> . How to reverse all the bits in a unsigned integer? an signed
> integer (you have to keep the sign of the integer)?
For an unsigned int, the following algorithm can be used (I believe
you can find in in the "Hacker's Delight" book by H.War
There is a highly optimal O(n) solution available.
1) Find the depth at which each node is situated let them be du and dv
respectively for u and v
2) Not bring u and v to the same depth. that is skip abs(du-dv) nodes from
the greater depth node
3) Compare both nodes and do a (u=u->parent and v=v->
Find the longest common prefix in the path from the root to nodes u and v
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The standard link
http://graphics.stanford.edu/~seander/bithacks.html
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int *p;
void PrintallComb(int i)
{
int j=0;
if(i == nBits)
{
for(j=0;jhttp://groups.google.com/group/algogeeks
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I have a simple method.
Let hash[x]=count of 1 bits in x for 0<=x<2^16
So for any 32-bit integer x, the count of 1 bits is hash[x&0x]
+hash[x>>16].
We can pre-calculate all hash[] value.
On Mar 21, 1:24 pm, "hijkl" <[EMAIL PROTECTED]> wrote:
> How to count the 1 bits in a integer?
> int C
Hi,,
Think the No of 1 bits can be found in o(k) where k is the no of bits
set
while((n & (n-1)) >= 0)
{
c++;
n = n & (n-1);
printf(">> n %d Cnt %d \n",n,c);
if(!n)
break;
}
printf("Cnt %d \n",c);
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