在二进制搜索和顺序搜索时比较计数比较麻烦，需要清楚

Question

I have a quick quick question but I am having trouble understanding, code sample below with commented documentation.

When searching through an array of elements using binarySearch what is considering a comparison?

or in the case of int array containing 178,384,471,478,487,576,617,630,649,817,821,915,917,972,975 with 821 being the target what would the comparison count be???

I have posted where I think comparisons take place, please correct me if I'm wrong, or tell me if I'm right please :) thanks all.

public static int binarySearch(String[] data, String target, int first, int last)
{
  if(first > last)
     return -1;
  else
    int middle = (first+last)/2;
    int result = data[first].compareTo(data[middle])
    if(result == 0) //IS THIS A COMPARISON? (I THINK YES)
      return middle;
    else if(result < 0) // IS THIS A COMPARISON (I THINK YES)
      return binarySearch(data,target.first,middle-1);
    else
      return binarySearch(data,target,middle+1,last); // IS THIS A COMPARISON? (I THINK NO)
}

Answer 1

实际上，这是真实的比较int result = data[first].compareTo(data[middle]) if语句仅基于“真实”比较做不同的事情。

Answer 2

So, almost by definition, any equality or relational operator or method that wraps such an operator would be a comparison.

In general, what you probably want to assess is the order of your algorithms as the input grows, it is enough to count say times that the function gets called, even if inside there are 3 comparisons for instance. The order (in this case O(log(n)) will not change.

Side notes: you might want to use a actual array of integers instead of string comparison, which can be costly AND tricky (ie "12" > "110"). Also, it would eliminate the need to get a "result", just do comparison against the stored values.