已排序数组的二分查找

Question

我正在尝试使用此二进制搜索代码搜索降序排序的数组。然而，在我对它进行排序并尝试搜索之后，它没有返回任何结果，只是一个加载图标，它永远不会消失，就好像它有一个无限循环一样。我不确定问题是什么，因为代码看起来合乎逻辑。

这是4.0框架的aspx，c#。提前致谢！

    protected void Button2_Click(object sender, EventArgs e)
    {
        String item = TextBox1.Text;
        int target = Convert.ToInt16(item);
        int mid, first = 0, last = mynumbers.Length - 1;

        //for a sorted array with descending values
        while (first<=last)
        {
            mid = (first + last) / 2;
            if (target < mynumbers[mid])
                first = mid + 1;
            if (target > mynumbers[mid])
                last = mid - 1;
            else
                Label11.Text = "Target " + item + " was found at index " + mynumbers[mid];

        }

Answer 1

Array 类中有一个二分搜索：

int index = Array.BinarySearch(mynumbers, target);

对于降序排列，可以使用 ReverseComparer 轻松完成，它很容易编写如下：

    public class ReverseComparer<T> : IComparer<T>
    {
        public int Compare(T x, T y)
        {
            return Comparer<T>.Default.Compare(y, x);
        }
    }

那么：

int index = Array.BinarySearch(numbers, 7, new ReverseComparer<int>());

如果这是一个学术问题练习时，您必须使用自定义搜索，当然，这不适用。如果它必须是类的自定义算法，那么问题是您必须在找到时跳出循环，并且索引位于 mid，而不是 mynumbers[mid]：

    //for a sorted array with descending values
    while (first<=last)
    {
        mid = (first + last) / 2;

        if (target < mynumbers[mid])
        {
            first = mid + 1;
        }

        if (target > mynumbers[mid])
        {
            last = mid - 1;
        }

        else
        {
            // the index is mid, not mynumbers[mid], and you need to break here
            // once found or it's an infinite loop once it finds it.
            Label11.Text = "Target " + item + " was found at index " + mid;
            break;
        }
    }

实际上，我可能会设置一个 bool 标志来保持算法的纯粹性，而不是将查找与输出问题混合在一起，这也将使您更容易知道如果退出循环而未找到会发生什么:

    bool found = false;

    //for a sorted array with descending values
    while (!found && first<=last)
    {
        mid = (first + last) / 2;

        if (target < mynumbers[mid])
        {
            first = mid + 1;
        }

        if (target > mynumbers[mid])
        {
            last = mid - 1;
        }

        else
        {
            // You need to stop here once found or it's an infinite loop once it finds it.
            found = true;
        }
    }

    Label11.Text = found 
        ? "Item " + item + " was found at position " + mid
        : "Item " + item + " was not found";

Answer 2

在黑暗中拍摄：

if (target < mynumbers[mid]) 
   first = mid + 1; 
else if (target > mynumbers[mid]) 
   last = mid - 1; 
else 
{
    ....
    break;
}

我怀疑你在 mid+1 和 mid-1 之间来回跳动

Answer 3

这是一个正确的：

如果目标< mynumbers[mid] 那么你必须把最后一个带到 mid-1，
因为我们必须在较低范围内搜索，即从第一个到中间 1

while (first<=last)
        {
            mid = (first + last) / 2;
            if (target == mynumbers[mid])
            Label11.Text = "Target " + item + " was found at index " + mynumbers[mid];

            else if (target < mynumbers[mid])
                last = mid - 1;
            else (target > mynumbers[mid])
                first = mid + 1;

            }

Answer 4

//this works fine with these Test cases    
// has to check if (target == mynumbers[mid])    
// this is for an array with ascending order.
class Program
{

    static void Main(string[] args)
    {
        // TEST cases:
        // for 8: item 8 was not found
        // for 4: item 4 found at Position 3
        // for 1: item 1 found at position 0
        // for 0: item 0 was not found


        int target =8;
        int searchkey = target;

        int[] mynumbers = { 1, 2, 3, 4, 5 };

        int mid=0, first = 0, last = mynumbers.Length - 1;

        bool found = false;

        //for a sorted array with ascending values
        while (!found && first <= last)
        {
            mid = (first + last) / 2;

            if (target == mynumbers[mid])
                found = true;
            else
            {

                if (target > mynumbers[mid])
                {
                    first = mid + 1;
                }

                if (target < mynumbers[mid])
                {
                    last = mid - 1;
                }

            }

        }

        String foundmsg = found
            ? "Item " + searchkey + " was found at position " + mid
            : "Item " + searchkey + " was not found";
        Console.WriteLine(foundmsg);
     }
}

Answer 5

它对我有用

public static int search(int[] arr, int value)
{
    Debug.Assert(arr.Length>0);
    var left = 0;
    var right = arr.Length - 1;

    while (((right - left)/2) > 0)
    {
        var middle = left + (right - left)/2;
        if (arr[middle] < value)
            left = middle + 1;
        else
            right = middle;
    }
    return arr[left] >= value ? left : right;
}

Answer 6

在 .NET 8 中使用 C# 12 进行二分搜索。

public static class BinarySearcher
{
    public static int FindIndex(int[] sortedData, int item)
    {
        // leftIndex and rightIndex keep track of which part of the array you’ll search in.
        var leftIndex = 0;
        var rightIndex = sortedData.Length - 1;

        // While you haven’t narrowed it down to one element
        while (leftIndex <= rightIndex)
        {
            // Check the middle element
            var midIndex = (leftIndex + rightIndex) / 2;
            var guess = sortedData[midIndex];
            
            if (guess == item) return midIndex;

            // The guess was too high
            if (guess > item)
            {
                rightIndex = midIndex - 1;
            }
            // The guess was too low
            else
            {
                leftIndex = midIndex + 1;
            }
        }
        
        return -1; // The item does not exist.
    }
}

使用 NUnit 3 进行测试。

using NUnit.Framework;

public class BinarySearcherTests
{
    [Test]
    public void BinarySearcher_Finds_Item()
    {
        // Arrange
        int[] myList = [1, 3, 5, 7, 9];
        var myListItem = myList[3];
        // Act
        var searchedItemIndex = BinarySearcher.FindIndex(myList, myListItem);
        // Assert
        Assert.That(searchedItemIndex, Is.EqualTo(3));
    }

    [Test]
    public void BinarySearcher_Doesnt_Find_Item()
    {
        // Arrange
        int[] myList = [1, 3, 5, 7, 9];
        // Act
        var searchedItemIndex = BinarySearcher.FindIndex(myList, 10);
        // Assert
        Assert.That(searchedItemIndex, Is.EqualTo(-1));
    }
}

Answer 7

public static object BinnarySearch(int[] array,int SearchKey)
    {
        int min = 0;
        int max = array.Length - 1;
        while (min <= max)
        {
            int mid = (min + max) / 2;
            if (SearchKey == array[mid])
            {
                return mid;
            }
            else if (SearchKey < array[mid])
            {
                max = mid - 1;
            }
            else
                min = mid + 1;
        }
        return "Not Found";
    }