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。
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):
RangeGroup
RangeGroup.Low
RangeGroups
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。