一尘不染

在范围列表中搜索数字的最快方法

algorithm

我有以下代码来查找范围列表中数字的匹配项。

public class RangeGroup
{
    public uint RangeGroupId { get; set; }
    public uint Low { get; set; }
    public uint High { get; set; }
    // More properties related with the range here
}

public class RangeGroupFinder
{
    private static readonly List<RangeGroup> RangeGroups=new List<RangeGroup>();

    static RangeGroupFinder()
    {
        // Populating the list items here
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023238144, High = 1023246335 });
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023246336, High = 1023279103 });
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023279104, High = 1023311871 });
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023311872, High = 1023328255 });
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023328256, High = 1023344639 });
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023344640, High = 1023410175 });
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023410176, High = 1023672319 });
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023672320, High = 1023688703 });
        RangeGroups.Add(new RangeGroup { RangeGroupId = 0, Low = 1023692800, High = 1023696895 });
       // There are many more and the groups are not sequential as it can seen on last 2 groups
    }

    public static RangeGroup Find(uint number)
    {
        return RangeGroups.FirstOrDefault(rg => number >= rg.Low && number <= rg.High);
    }
}

RangeGroup的列表包含大约5000000个项目,并且将大量使用Find()方法,因此我正在寻找一种进行搜索的更快方法。更改数据结构或以任何方式拆分数据都没有问题。

编辑:

所有范围都是唯一的,并且按“低”的顺序添加,并且它们不重叠。

结果:

使用ikh的代码进行了测试,结果比我的代码快大约7000倍。测试代码和结果可以在这里看到。


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2020-07-28

共1个答案

一尘不染

由于您指出RangeGroups是按的顺序添加的,RangeGroup.Low并且它们不重叠,因此您无需进行任何进一步的预处理。您可以在RangeGroups列表上进行二进制搜索以找到范围(警告:未经充分测试,您需要检查一些边缘条件):

public static RangeGroup Find(uint number) {
    int position = RangeGroups.Count / 2;
    int stepSize = position / 2;

    while (true) {
        if (stepSize == 0) {
            // Couldn't find it.
            return null;
        }

        if (RangeGroups[position].High < number) {
            // Search down.
            position -= stepSize;

        } else if (RangeGroups[position].Low > number) {
            // Search up.
            position += stepSize;

        } else {
            // Found it!
            return RangeGroups[position];
        }

        stepSize /= 2;
    }
}

最坏情况下的运行时间应该在O(log(N))左右,其中N是RangeGroups的数量。

2020-07-28