Question

假设我有一个10个整数的数组，我正在使用二进制搜索来查找数字，让我们以数字为例

1 2 3 4 5 6 7 8 9 10

我正在使用这种方法

static void binarySearch(int n, int[] a, int low, int high)
    {
        int mid = (high + low) / 2;
        if(low > high)
            System.out.println(n+" was not found after "+counter+" comparisons");
        else if(a[mid] == n)
        {
            counter++;
            System.out.println(n+" was found at position "+mid+" after "+counter+" comparisons");
        }            
        else if(a[mid] < n)
        {
            counter++;
            binarySearch(n, a, mid+1, high);
        }            
        else
        {
            counter++;
            binarySearch(n, a, low, mid-1);
        }            
    }

调用方法binarySearch（5，a，0，a.lenght）的正确方法是什么要么 binarySearch（5，a，0，a.lenght-1）

我知道他们都会找到这个号码，但他们会在不同的索引中找到它;从而进行更多的比较

Answer 1

正确的方法是避免使用这种方法，并使用标准Arrays.binarySearch()方法，该方法具有记录的巨大优势，以及返回结果的另一个巨大优势，而不是在System.out上打印它（这使它无用）。

Answer 2

我们应该做一些测试吗？

首先，让我们搜索数组中的每个数字。我们得到：

binarySearch(i, array, 0, array.length);

1 was found at position 0 after 3 comparisons
2 was found at position 1 after 4 comparisons
3 was found at position 2 after 2 comparisons
4 was found at position 3 after 3 comparisons
5 was found at position 4 after 4 comparisons
6 was found at position 5 after 1 comparisons
7 was found at position 6 after 3 comparisons
8 was found at position 7 after 4 comparisons
9 was found at position 8 after 2 comparisons
10 was found at position 9 after 3 comparisons
Average: 2.9 comparisons

binarySearch(i, array, 0, array.length - 1);

1 was found at position 0 after 3 comparisons
2 was found at position 1 after 2 comparisons
3 was found at position 2 after 3 comparisons
4 was found at position 3 after 4 comparisons
5 was found at position 4 after 1 comparisons
6 was found at position 5 after 3 comparisons
7 was found at position 6 after 4 comparisons
8 was found at position 7 after 2 comparisons
9 was found at position 8 after 3 comparisons
10 was found at position 9 after 4 comparisons
Average: 2.9 comparisons

如您所见，差异确实会出现，但平均值保持不变。现在让我们测试更大的数字：

100000 items
binarySearch(i, array, 0, array.length);
Average: 15.68946 comparisons
binarySearch(i, array, 0, array.length - 1);
Average: 15.68946 comparisons

200000 items
binarySearch(i, array, 0, array.length);
Average: 16.689375 comparisons
binarySearch(i, array, 0, array.length - 1);
Average: 16.689375 comparisons

500000 items
binarySearch(i, array, 0, array.length);
Average: 17.951464 comparisons
binarySearch(i, array, 0, array.length - 1);
Average: 17.951464 comparisons

因此，平均而言，它不会走向任何一种方式。为了惯例，我建议使用独有的上限版本：binarySearch(i, array, 0, array.length);

Answer 3

可以提出这样的问题：正确的包容性 [低，高] 还是正确的 [低，高] ？这种独特的形式具有由Dijkstra“创立”的悠久计算机科学传统。在我看来，它也更优雅（a.length而不是a.length-1）。

但是在你的函数中它是[低，高]，a.length-1，因为你看到低＆gt;高（不低> =高）和（低，中1）（不是（低，中））。

调用二进制搜索方法的正确方法是什么

3 个答案: