我正在寻找一个关于哈希表如何工作的解释-用简单的英语为我这样的傻瓜!

例如,我知道它接受键,计算哈希(我正在寻找一个解释),然后执行某种模运算来计算出它在存储值的数组中的位置,但我的知识到此为止。

谁能解释一下过程吗?

编辑:我并不是特别问哈希码是如何计算的,而是对哈希表如何工作的一般概述。


当前回答

哈希表完全基于这样一个事实,即实际计算遵循随机访问机模型,即内存中任何地址的值都可以在O(1)时间或常数时间内访问。

因此,如果我有一个键的宇宙(我可以在应用程序中使用的所有可能的键的集合,例如,滚动no。对于学生来说,如果它是4位,那么这个宇宙就是从1到9999的一组数字),并且一种将它们映射到有限大小的数字集的方法可以在我的系统中分配内存,理论上我的哈希表已经准备好了。

Generally, in applications the size of universe of keys is very large than number of elements I want to add to the hash table(I don't wanna waste a 1 GB memory to hash ,say, 10000 or 100000 integer values because they are 32 bit long in binary reprsentaion). So, we use this hashing. It's sort of a mixing kind of "mathematical" operation, which maps my large universe to a small set of values that I can accomodate in memory. In practical cases, often space of a hash table is of the same "order"(big-O) as the (number of elements *size of each element), So, we don't waste much memory.

现在,一个大集合映射到一个小集合,映射必须是多对一的。因此,不同的键将被分配相同的空间(?? ?不公平)。有几种方法可以解决这个问题,我只知道其中最流行的两种:

Use the space that was to be allocated to the value as a reference to a linked list. This linked list will store one or more values, that come to reside in same slot in many to one mapping. The linked list also contains keys to help someone who comes searching. It's like many people in same apartment, when a delivery-man comes, he goes to the room and asks specifically for the guy. Use a double hash function in an array which gives the same sequence of values every time rather than a single value. When I go to store a value, I see whether the required memory location is free or occupied. If it's free, I can store my value there, if it's occupied I take next value from the sequence and so on until I find a free location and I store my value there. When searching or retreiving the value, I go back on same path as given by the sequence and at each location ask for the vaue if it's there until I find it or search all possible locations in the array.

CLRS的《算法导论》对这个主题提供了非常好的见解。

其他回答

这是另一种看待它的方式。

我假设你理解数组A的概念,它支持索引操作,你可以一步找到第I个元素,A[I],不管A有多大。

因此,例如,如果您想存储一组恰好年龄不同的人的信息,一个简单的方法是有一个足够大的数组,并使用每个人的年龄作为数组的索引。这样,你就可以一步获取任何人的信息。

But of course there could be more than one person with the same age, so what you put in the array at each entry is a list of all the people who have that age. So you can get to an individual person's information in one step plus a little bit of search in that list (called a "bucket"). It only slows down if there are so many people that the buckets get big. Then you need a larger array, and some other way to get more identifying information about the person, like the first few letters of their surname, instead of using age.

这是基本思想。不使用年龄,可以使用任何能产生良好价值观传播的人的函数。这就是哈希函数。比如你可以把这个人名字的ASCII表示的每三分之一,按某种顺序打乱。重要的是,您不希望太多人散列到同一个存储桶,因为速度取决于存储桶保持较小。

直连地址表

要理解哈希表,直接地址表是我们应该理解的第一个概念。

直接地址表直接使用键作为数组中槽的索引。宇宙键的大小等于数组的大小。在O(1)时间内访问这个键非常快,因为数组支持随机访问操作。

然而,在实现直接地址表之前,有四个注意事项:

要成为有效的数组索引,键应该是整数 键的范围是相当小的,否则,我们将需要一个巨大的数组。 不能将两个不同的键映射到数组中的同一个槽 宇宙键的长度等于数组的长度

事实上,现实生活中并不是很多情况都符合上述要求,所以哈希表就可以救场了

哈希表

哈希表不是直接使用键,而是首先应用数学哈希函数将任意键数据一致地转换为数字,然后使用该哈希结果作为键。

宇宙键的长度可以大于数组的长度,这意味着两个不同的键可以散列到相同的索引(称为散列碰撞)?

实际上,有一些不同的策略来处理它。这里有一个常见的解决方案:我们不将实际值存储在数组中,而是存储一个指向链表的指针,该链表包含散列到该索引的所有键的值。

如果你仍然有兴趣知道如何从头开始实现hashmap,请阅读下面的帖子

用法和行话:

哈希表用于快速存储和检索数据(或记录)。 记录使用散列键存储在桶中 哈希键是通过对记录中包含的选定值(键值)应用哈希算法来计算的。所选值必须是所有记录的公共值。 每个桶可以有多条记录,这些记录按照特定的顺序组织。

现实世界的例子:

哈希公司成立于1803年,当时没有任何计算机技术,只有300个文件柜来保存大约3万名客户的详细信息(记录)。每个文件夹都清楚地标识其客户端编号,从0到29,999的唯一编号。

当时的档案管理员必须迅速为工作人员获取和存储客户记录。工作人员决定使用哈希方法来存储和检索他们的记录会更有效。

要归档客户记录,档案管理员将使用写在文件夹上的唯一客户编号。使用这个客户端编号,他们将哈希键调整300,以识别包含它的文件柜。当他们打开文件柜时,他们会发现里面有很多按客户号排序的文件夹。在确定正确的位置后,他们会简单地把它塞进去。

要检索客户记录,档案管理员将在一张纸上获得客户号码。使用这个唯一的客户端编号(哈希键),他们会将其调整300,以确定哪个文件柜拥有客户端文件夹。当他们打开文件柜时,他们会发现里面有很多按客户号排序的文件夹。通过搜索记录,他们可以快速找到客户端文件夹并检索它。

在我们的实际示例中,桶是文件柜,记录是文件夹。


需要记住的一件重要的事情是,计算机(及其算法)处理数字比处理字符串更好。因此,使用索引访问大型数组要比按顺序访问快得多。

正如Simon提到的,我认为非常重要的是哈希部分是转换一个大空间(任意长度,通常是字符串等),并将其映射到一个小空间(已知大小,通常是数字)进行索引。记住这一点非常重要!

因此,在上面的示例中,大约30,000个可能的客户机被映射到一个较小的空间中。


这样做的主要思想是将整个数据集划分为几个部分,以加快实际搜索的速度,而实际搜索通常是耗时的。在我们上面的例子中,300个文件柜中的每个(统计上)将包含大约100条记录。搜索100条记录(不管顺序)要比处理3万条记录快得多。

你可能已经注意到有些人已经这样做了。但是,在大多数情况下,他们只是使用姓氏的第一个字母,而不是设计一个哈希方法来生成哈希键。因此,如果您有26个文件柜,每个文件柜都包含从a到Z的一个字母,理论上您只是将数据分割并增强了归档和检索过程。

其实比这更简单。

哈希表不过是一个包含键/值对的向量数组(通常是稀疏数组)。此数组的最大大小通常小于哈希表中存储的数据类型的可能值集中的项数。

哈希算法用于根据将存储在数组中的项的值生成该数组的索引。

This is where storing vectors of key/value pairs in the array come in. Because the set of values that can be indexes in the array is typically smaller than the number of all possible values that the type can have, it is possible that your hash algorithm is going to generate the same value for two separate keys. A good hash algorithm will prevent this as much as possible (which is why it is relegated to the type usually because it has specific information which a general hash algorithm can't possibly know), but it's impossible to prevent.

因此,您可以使用多个键来生成相同的散列代码。当这种情况发生时,将遍历向量中的项,并在向量中的键和正在查找的键之间进行直接比较。如果找到,则返回与该键关联的值,否则不返回任何值。

有很多答案,但没有一个是非常可视化的,而哈希表在可视化时很容易“点击”。

哈希表通常实现为链表数组。如果我们想象一个存储人名的表,经过几次插入之后,它可能会被放置在内存中,其中()包含的数字是文本/姓名的哈希值。

bucket#  bucket content / linked list

[0]      --> "sue"(780) --> null
[1]      null
[2]      --> "fred"(42) --> "bill"(9282) --> "jane"(42) --> null
[3]      --> "mary"(73) --> null
[4]      null
[5]      --> "masayuki"(75) --> "sarwar"(105) --> null
[6]      --> "margaret"(2626) --> null
[7]      null
[8]      --> "bob"(308) --> null
[9]      null

以下几点:

each of the array entries (indices [0], [1]...) is known as a bucket, and starts a - possibly empty - linked list of values (aka elements, in this example - people's names) each value (e.g. "fred" with hash 42) is linked from bucket [hash % number_of_buckets] e.g. 42 % 10 == [2]; % is the modulo operator - the remainder when divided by the number of buckets multiple data values may collide at and be linked from the same bucket, most often because their hash values collide after the modulo operation (e.g. 42 % 10 == [2], and 9282 % 10 == [2]), but occasionally because the hash values are the same (e.g. "fred" and "jane" both shown with hash 42 above) most hash tables handle collisions - with slightly reduced performance but no functional confusion - by comparing the full value (here text) of a value being sought or inserted to each value already in the linked list at the hashed-to bucket

链表长度与负载因子有关,而不是值的数量

如果表的大小增加,上面实现的哈希表倾向于调整自己的大小(即创建一个更大的桶数组,在那里创建新的/更新的链表,删除旧的数组),以保持值与桶的比率(又名负载因子)在0.5到1.0的范围内。

Hans gives the actual formula for other load factors in a comment below, but for indicative values: with load factor 1 and a cryptographic strength hash function, 1/e (~36.8%) of buckets will tend to be empty, another 1/e (~36.8%) have one element, 1/(2e) or ~18.4% two elements, 1/(3!e) about 6.1% three elements, 1/(4!e) or ~1.5% four elements, 1/(5!e) ~.3% have five etc.. - the average chain length from non-empty buckets is ~1.58 no matter how many elements are in the table (i.e. whether there are 100 elements and 100 buckets, or 100 million elements and 100 million buckets), which is why we say lookup/insert/erase are O(1) constant time operations.

哈希表如何将键与值关联

Given a hash table implementation as described above, we can imagine creating a value type such as `struct Value { string name; int age; };`, and equality comparison and hash functions that only look at the `name` field (ignoring age), and then something wonderful happens: we can store `Value` records like `{"sue", 63}` in the table, then later search for "sue" without knowing her age, find the stored value and recover or even update her age - happy birthday Sue - which interestingly doesn't change the hash value so doesn't require that we move Sue's record to another bucket.

当我们这样做的时候,我们使用哈希表作为一个关联容器,也就是map,它存储的值可以被认为是由一个键(名称)和一个或多个其他字段组成,仍然被称为值(在我的例子中,只是年龄)。用作映射的哈希表实现称为哈希映射。

这与前面我们存储离散值的例子形成了对比,比如“sue”,你可以把它看作是它自己的键:这种用法被称为散列集。

还有其他方法来实现哈希表

并不是所有的哈希表都使用链表(称为独立链表),但大多数通用哈希表都使用链表,因为主要的替代封闭哈希(又名开放寻址)-特别是支持擦除操作-与易于冲突的键/哈希函数相比性能不太稳定。


简单讲一下哈希函数

强大的散列…

一个通用的、最小化最坏情况碰撞的哈希函数的工作是有效地随机地在哈希表桶周围散布键,同时总是为相同的键生成相同的哈希值。理想情况下,即使在键的任何位置改变一个位,也会随机地翻转结果哈希值中的大约一半位。

This is normally orchestrated with maths too complicated for me to grok. I'll mention one easy-to-understand way - not the most scalable or cache friendly but inherently elegant (like encryption with a one-time pad!) - as I think it helps drive home the desirable qualities mentioned above. Say you were hashing 64-bit doubles - you could create 8 tables each of 256 random numbers (code below), then use each 8-bit/1-byte slice of the double's memory representation to index into a different table, XORing the random numbers you look up. With this approach, it's easy to see that a bit (in the binary digit sense) changing anywhere in the double results in a different random number being looked up in one of the tables, and a totally uncorrelated final value.

// note caveats above: cache unfriendly (SLOW) but strong hashing...
std::size_t random[8][256] = { ...random data... };
auto p = (const std::byte*)&my_double;
size_t hash = random[0][p[0]] ^
              random[1][p[1]] ^
              ... ^
              random[7][p[7]];

弱但通常快速的哈希…

Many libraries' hashing functions pass integers through unchanged (known as a trivial or identity hash function); it's the other extreme from the strong hashing described above. An identity hash is extremely collision prone in the worst cases, but the hope is that in the fairly common case of integer keys that tend to be incrementing (perhaps with some gaps), they'll map into successive buckets leaving fewer empty than random hashing leaves (our ~36.8% at load factor 1 mentioned earlier), thereby having fewer collisions and fewer longer linked lists of colliding elements than is achieved by random mappings. It's also great to save the time it takes to generate a strong hash, and if keys are looked up in order they'll be found in buckets nearby in memory, improving cache hits. When the keys don't increment nicely, the hope is they'll be random enough they won't need a strong hash function to totally randomise their placement into buckets.