没有你的车你就没有那么坚强。最快的查找列表

发布于 2024-08-08 00:53:04 字数 445 浏览 9 评论 0原文

我有一个结构集合。每个结构有 2 个键。如果我使用键 #1 查询,我应该得到键 #2 作为回报,反之亦然。

当您拥有 .NET Framework 的强大支持时,可以轻松地在桌面上编写代码。我正在 .NET Micro Framework 中编写代码,这是一个非常非常有限的框架子集。例如,就集合而言,我只有数组和 ArrayList 对象可供使用。

例如,这里是结构列表:

Key #1        Key #2 
6             A
7             F
8             Z
9             B

因此,当我查询 8 时,我应该得到 Z。 当我查询 Z 时,我应该得到 8。

我希望使用数组或 ArrayList 进行最快且处理器密集程度最低的查找。我正在编码的设备是低端 ARM 处理器,因此我需要尽早优化。

I have a collection of structures. Each structure has 2 keys. If I query using key #1, I should get key #2 in return and vice versa.

It's easy to write code on the desktop when you have the power of the .NET Framework behind you. I am writing code in the .NET Micro Framework, which is a very very limited subset of framework. For instance, as far as collections, I only have arrays and ArrayList objects at my disposal.

So for example here is the list of structures:

Key #1        Key #2 
6             A
7             F
8             Z
9             B

So when I query for 8, I should get Z.
When I query for Z, I should get 8.

I am looking to do the fastest and least processor intensive lookup using either arrays or ArrayList. The device I am coding against is a low-end ARM processor, thus I need to optimize early.

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评论(8

弄潮 2024-08-15 00:53:04

如果该集合是固定的,请查看完美哈希函数

If the set is fixed, look into perfect hash functions.

烟酉 2024-08-15 00:53:04

你有什么理由不能编写自己的哈希图吗?

Any reason you can't write your own hashmap?

冷︶言冷语的世界 2024-08-15 00:53:04

这取决于条目数量和您的访问模式。

鉴于您的访问模式是随机访问,如果您没有太多元素,您可以拥有 2 个对数组,然后使用

Array.BinarySearch()

It depends on the number of entries and your access pattern.

Given that your access pattern is random access if you don't have too many elements you could have 2 Arrays of Pairs and then use

Array.BinarySearch()
迷荒 2024-08-15 00:53:04

好吧...如果您想要最快并且不太关心内存,只需使用两个哈希表即可。一个人往一个方向走,一个人往另一个方向走。不确定是否有一种更有效的内存方式...

或者只使用一个哈希表,但其中有两个方向的条目。

Well... if you want the fastest and aren't too concerned about memory, just use two hash tables. One going one way, one going to other. Not sure if there's a more memory efficient way...

Or use just one hash table but have the entries for both directions in there.

幸福%小乖 2024-08-15 00:53:04

它不是那么简单吗:

  • 在您正在查询的数组中找到键,
  • 返回相反数组中相同索引处的键

我会尽可能保持简单,只需迭代您的数组正在寻找。如果您的列表(从空气中提取数字)超过 1k+ 元素,您可能只会看到实现一些散列例程的好处,并且您自己的散列例程增加的复杂性会在一定程度上减慢速度。

Is it not as simple as :

  • find the key in the array you're querying
  • return the key at the same index in the opposite array

I would keep it as simple as possible and just iterate through the array you're searching. You'll probably only see a benefit from implementing some hashing routines if your list is (plucks figure from air) over 1k+ elements, with the added complexity of your own hashing routines slowing things down somewhat.

黑寡妇 2024-08-15 00:53:04

几种解决方案:

  1. 保持两个列表同步,进行线性搜索。除非您的集合非常大,或者您正在重复搜索,否则效果很好。
  2. 两个哈希表。编写自己的相当容易——它只是一个固定的存储桶数组(每个存储桶可以是一个 ArrayList)。通过执行 object.GetHashCode() % numBuckets 将项目映射到存储桶。
  3. 两个数组的大小值的范围。如果您的数字在固定范围内,请分配一个该范围大小的数组,其中元素是来自另一组的项目。超级快速和简单,但占用大量内存。

Several solutions:

  1. Keep 2 lists in sync, do a linear search. Works well unless your collections are very large, or you're searching repeatedly.
  2. Two hashtables. Writing your own is fairly easy -- it is just a fixed array of buckets (each bucket can be an ArrayList). Map an item to a bucket by doing object.GetHashCode() % numBuckets.
  3. Two arrays the size of the range of values. If your numbers are in a fixed range, allocate an array the size of the range, with elements being items from the other group. Super quick and easy, but uses a lot of memory.
蘸点软妹酱 2024-08-15 00:53:04

如果是固定集,请考虑使用switch。请参阅此处类似问题的答案

If it's a fixed set, consider using switch. See the answer to a similar question here.

终止放荡 2024-08-15 00:53:04

几年前,我在编写 C 语言时遇到了这个问题,我们需要在大约 10,000 行中快速找到条形码(数字)(当时,使用文件作为数据库 - 因为它是一个手持设备)

我创建了自己的搜索总是从中间开始,而不是逐一迭代...

在 10000 个项目堆栈中搜索 4050

从 5000 ( 10 000 / 2 ) 开始
现在,数字是更高还是更低...更低

从 2500 ( 5000 / 2 ) 开始
现在,数字是更高还是更低......更高

从 3750 ( 2500 + 2500 / 2 ) 开始
现在,数字是更高还是更低......更高

从 4375 ( 3750 + 1250 / 2 ) 开始
现在,数字是更高还是更低……更低

从 4063 ( 4375 - 625 / 2 ) 开始
现在,数字是更高还是更低...更低

从 3907 ( 4063 - 312 / 2 ) 开始
现在,数字是更高还是更低……更高

从 3907 开始 ( 3907 + 156 / 2 )
现在,数字是更高还是更低……更高

从 3946 ( 3907 + 78 / 2 ) 开始
现在,数字是更高还是更低……更高

...

直到获得值...您将需要搜索大约 14 次,而不是

关于字母的 4050 次迭代...它们也都代表一个数字.. 。

希望有帮助

I had this problem several years ago when programming C, that we need to find a barcode (numeric) quickly in about 10 thousand rows (in that time, using a file as the Database - as it was a hand device)

I created my own search that instead of iterate one by one would start always in the middle...

searching for 4050 in 10000 item stack

start on 5000 ( 10 000 / 2 )
now, is the number higher or lower ... lower

start on 2500 ( 5000 / 2 )
now, is the number higher or lower ... higher

start on 3750 ( 2500 + 2500 / 2 )
now, is the number higher or lower ... higher

start on 4375 ( 3750 + 1250 / 2 )
now, is the number higher or lower ... lower

start on 4063 ( 4375 - 625 / 2 )
now, is the number higher or lower ... lower

start on 3907 ( 4063 - 312 / 2 )
now, is the number higher or lower ... higher

start on 3907 ( 3907 + 156 / 2 )
now, is the number higher or lower ... higher

start on 3946 ( 3907 + 78 / 2 )
now, is the number higher or lower ... higher

...

until you get the value... you will need to search about 14 times instead 4050 iterations

about the letters ... they all represent a numeric number as well...

Hope it helps

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