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Why take binary search runtime complexity into account - log2N

Can someone explain to me when it comes to the “Binary” search, we say that the runtime complexity is O (log n)? I searched it on google and got below

"The number of times you can halve your search space is the same as log 2 n ".

I know that we occupy half until we find the search key in the data structure, but why should we treat it as a log 2n? I understand that e x is exponential growth, so log 2n is binary decay. But I cannot interpret binary search in terms of understanding the definition of a logarithm.

thank

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2 answers

:

- m (.. m), ?

, 2 m right?

, n = 2 m , , m, m n :

n = 2 m

log 2 (n) = log 2 (2 m)

log 2 (n) = m


: n = 2 m , m, , , log 2 (n).

+13

: -

.

, "n" , . , "1".

n : a a a a.... ( n)

: a a a..a( n/2)

: a a a.. a ( n/(2 ^ 2));

onn.............. , k 1
i.e n/(2 ^ k) 1 atlast

/(2 ^ ) = 1; 2 ^ = ; log 2

k = log n;

-, n n/2 n/4 n/8 , - log n

+1

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