二元搜索的比较次数

Question

What is the total number of comparisons necessary to locate all the n sorted distinct integers in an array using binary search? I think the number is n log ₂ n (2 is the base), but I am not sure. What do you think?

Answer 1

If you want an exact answer, then it is clearly not N log(N) or N log ₂ (N). For most integers N, logN and log ₂ are not rational, but the number of comparisons must be an integer value.

Also, the exact answer will depend on implementation details of the binary search algorithm. For example, if a "comparison" is a simple relation that returns true and false, more comparisons are required than when a "comparison" returns negative, zero or positive. (In the latter case, you can short circuit when the algorithm hits the key early.)

Answer 2

找到每一个需要log2 n ，它需要完成n次，所以n log n就是这样。

Answer 3

I would say it takes n log m

Where m is the size of the array, so log m to find a value. And then you do this n times for the n distinct integers.

So n log m in total.

Answer 4

There will be at most (2 * log ₂ n + 1) rounded down(so 7.6 => 7) comparisons for 1 number.

When we land on some number in array, first we check if it is the one we are looking for. (== first comparison). After that we check if it smaller (or greater) (second comparison).

To find the number, we must process at most log ₂ n numbers.

And we must make the last comparison on the last number, to check that this is the one.

So looking for 16 in [1..16] will take 2*log ₂ 16 + 1 = 9 comparisons (assuming we land on these numbers: 8, 12, 14, 15, 16). And looking for 10 in [1..10] will take 2*log ₂ 10 + 1 = 7.6 => 7 (Assuming we land on these numbers: 5, 8, 9, 10).

So for n numbers there will be at most n times more.

Answer 5

Thanks for your comments, now I am clear. I think what Stephen C said is true. I think we cannot actually work out a formula for this question unless we have a exact value. However, if the n=2^n -1,like 511(2^9 -1), it is easy to compute. For 511, total number of comparisons = 1x1 + 2X2 + 3X4 + 4X8 + 5X16 + 6X32 + 7X64 + 8X128 + 9X256 = 4097.