数组中数字的出现，具有复杂的log n算法和c

Question

嗨，天才，祝你有美好的一天 情况：

T [M] an array sorts in ascending order.

Write an algorithm that calculates the number of occurrences of an integer x in the array T[M].

the number of comparison of x to the elements of T, must be of order log n

我的尝试是：

def Occurrence(T,X,n):{
   
 
       if ( n=0 ){ return 0;}
        else { 

            Occurrence(T,X,n/2);

            if( T[n]==X ){  return 1+Occurrence(T,x,n/2); }
            else { return Occurrence(T,x,n/2); }

 
 }the end of code
complexity is :
                   0 if n=0
we have      O(n)={
                   1+O(n/2) if n>0
O(n)=1+1+1+....+O(n/2^(n))=n+O(2/2^(n))
when algorithm stopp if{existe k  n=2^(k),so   O(n)=n+1 } 
n/2^(n)=1)  =>   O(n)=log(n)+1, so you think my code is true ?
</pre>

Answer 1

查看 binary search variants 给出所需元素的最左边和最右边的索引，然后 return index_difference + 1

您可以在 C++ STL 中使用 lower_bound 和 upper_bound，在 Python 中使用 bisect_left 和 bisect_right，如果有类似的功能，或者从维基引用。

Answer 2

如果数组已排序

import bisect 

T = [1, 3, 4, 4, 4, 6, 7]
x = 4
right = bisect.bisect(T, x)
if(right == 0 or T[right - 1] != x):
    print("Count:0")
else:
    left = bisect.bisect_left(T, x)
    print("Count:", right - left)

2Log(n)

数组中数字的出现，具有复杂的log n算法和c

Occurrence of number in array,with complexity log n algorithm and c

algorithm

math

complexity-theory

division

find-occurrences