Gene wrote:
> Terry wrote:
> > If i have an array like {1,4, 10, 15 , 20 , 30 } of size n , now if i
> > want to search for number
> >
> > 25 , i should get 20 , if i search for number 11 i hould get 10 , if i
> > search for 4 i should get 4, if i search for a number and it doesn't
> > exist i should get the lower number between which it lies. All numbers
> > are bound to lie between 1 and n.
> >
> > Can you tell me a easy way to do it.
>
> If the array is always sorted you can modify a standard binary search
> to get job done in O(log n) time. This looks like homework.
Can someone give me a better version / elegant version
int
search (int low, int high, int num)
{
int mid = (high + low) / 2;
if (num <= arr1[0])
return 0;
if (num >= arr1[cnt - 1])
return cnt - 1;
if (arr1[mid] == num)
return mid;
if (arr1[high] == num)
return high;
if (arr1[low] == num)
return low;
if (low == high)
{
if (num == arr1[low])
return low;
if (num < arr1[low])
{
gflag = 1;
return low - 1;
}
if(num > arr1[low])
{
gflag=1;
return low;
}
}
if ((high - low) == 1)
{
gflag = 1;
return low;
}
if (arr1[mid] > num)
{
high = mid - 1;
return search (low, high, num);
}
if (arr1[mid] < num)
{
low = mid + 1;
return search (low, high, num);
}
}
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