哈希表是如何工作的?

这是一个外行的解释。

让我们假设你想要用书填满一个图书馆，而不仅仅是把它们塞进去，而且你希望在你需要它们的时候能够很容易地再次找到它们。

因此，您决定，如果想要阅读一本书的人知道书名和确切的书名，那么这就是所有应该做的。有了书名，在图书管理员的帮助下，读者就能轻松快速地找到这本书。

那么，你该怎么做呢?当然，你可以列出你把每本书放在哪里的列表，但是你会遇到和搜索图书馆一样的问题，你需要搜索列表。当然，列表会更小，更容易搜索，但您仍然不希望从库(或列表)的一端到另一端依次搜索。

你想要的东西，有了书名，就能立刻给你正确的位置，所以你所要做的就是漫步到正确的书架上，拿起书。

但这怎么能做到呢?嗯，当你填满图书馆的时候要有一点先见之明，当你填满图书馆的时候要做很多工作。

你设计了一个聪明的小方法，而不是开始从一端到另一端填满这个库。你拿着书名，在一个小的计算机程序中运行，它会显示出书架的编号和书架上的槽号。这是你放书的地方。

这个程序的美妙之处在于，稍后，当一个人回来阅读这本书时，您再次通过程序输入标题，并获得与最初给您的相同的书架编号和插槽编号，这就是书的位置。

正如其他人已经提到的，这个程序被称为哈希算法或哈希计算，通常通过输入数据(在这种情况下是书名)并从中计算一个数字来工作。

为了简单起见，我们假设它只是将每个字母和符号转换为一个数字，并将它们全部相加。实际上，它要比这复杂得多，但现在让我们先把它放在这里。

这种算法的美妙之处在于，如果你一次又一次地向它输入相同的输入，它每次都会输出相同的数字。

这就是哈希表的基本工作原理。

接下来是技术方面的内容。

首先是数字的大小。通常，这种哈希算法的输出在一个较大的数字范围内，通常比表中的空间大得多。例如，假设我们的图书馆刚好有100万本书的空间。哈希计算的输出可以在0到10亿的范围内，这要高得多。

那么，我们该怎么办呢?我们使用所谓的模量计算，它基本上是说，如果你数到你想要的数字(即10亿数字)，但想要保持在一个小得多的范围内，每次你达到这个小范围的极限，你就从0开始，但你必须跟踪你在大序列中走了多远。

假设哈希算法的输出在0到20的范围内，并且从特定的标题中获得值17。如果图书馆的大小只有7本书，你数1、2、3、4、5、6，当你数到7时，你从0开始。因为我们需要数17次，所以我们有1、2、3、4、5、6、0、1、2、3、4、5、6、0、1、2、3，最后的数字是3。

当然模量的计算不是这样的，它是用除法和余数来完成的。17除以7的余数是3(17除7得14,17和14之差是3)。

因此，你把书放在3号槽里。

这就导致了下一个问题。碰撞。由于该算法无法将图书间隔开来以使它们完全填满库(或者填满哈希表)，因此它最终总是会计算一个以前使用过的数字。在图书馆的意义上，当你到达书架和你想放一本书的槽号时，那里已经有一本书了。

存在各种冲突处理方法，包括将数据运行到另一个计算中以获得表中的另一个位置(双重哈希)，或者只是在给定的位置附近找到一个空间(例如，就在前一本书的旁边，假设插槽可用，也称为线性探测)。这意味着当你稍后试图找到这本书时，你需要做一些挖掘工作，但这仍然比简单地从图书馆的一端开始要好。

最后，在某些情况下，您可能希望将更多的书放入图书馆，而不是图书馆所允许的。换句话说，你需要建立一个更大的库。由于图书馆中的确切位置是使用图书馆的确切和当前大小计算出来的，因此，如果您调整了图书馆的大小，那么您可能最终不得不为所有书籍找到新的位置，因为为找到它们的位置所做的计算已经改变了。

我希望这个解释比桶和函数更接地气一点:)

2009-04-08 16:33:02

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。

2009-04-08 16:04:43

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.

2015-06-01 06:59:40

Hashtable inside contains cans in which it stores the key sets. The Hashtable uses the hashcode to decide to which the key pair should plan. The capacity to get the container area from Key's hashcode is known as hash work. In principle, a hash work is a capacity which when given a key, creates an address in the table. A hash work consistently returns a number for an item. Two equivalent items will consistently have a similar number while two inconsistent objects may not generally have various numbers. When we put objects into a hashtable then it is conceivable that various objects may have equal/ same hashcode. This is known as a collision. To determine collision, hashtable utilizes a variety of lists. The sets mapped to a single array index are stored in a list and then the list reference is stored in the index.

2021-11-30 15:01:54

你们已经很接近完整地解释了这个问题，但是遗漏了一些东西。哈希表只是一个数组。数组本身将在每个槽中包含一些内容。至少要将哈希值或值本身存储在这个插槽中。除此之外，您还可以存储在此插槽上碰撞的值的链接/链表，或者您可以使用开放寻址方法。您还可以存储一个或多个指针，这些指针指向您希望从该槽中检索的其他数据。

It's important to note that the hashvalue itself generally does not indicate the slot into which to place the value. For example, a hashvalue might be a negative integer value. Obviously a negative number cannot point to an array location. Additionally, hash values will tend to many times be larger numbers than the slots available. Thus another calculation needs to be performed by the hashtable itself to figure out which slot the value should go into. This is done with a modulus math operation like:

uint slotIndex = hashValue % hashTableSize;

这个值是该值将要进入的槽。在开放寻址中，如果槽位已经被另一个哈希值和/或其他数据填充，将再次运行模运算来查找下一个槽:

slotIndex = (remainder + 1) % hashTableSize;

我想可能还有其他更高级的方法来确定槽索引，但这是我见过的最常见的方法……会对其他表现更好的公司感兴趣。

With the modulus method, if you have a table of say size 1000, any hashvalue that is between 1 and 1000 will go into the corresponding slot. Any Negative values, and any values greater than 1000 will be potentially colliding slot values. The chances of that happening depend both on your hashing method, as well as how many total items you add to the hash table. Generally, it's best practice to make the size of the hashtable such that the total number of values added to it is only equal to about 70% of its size. If your hash function does a good job of even distribution, you will generally encounter very few to no bucket/slot collisions and it will perform very quickly for both lookup and write operations. If the total number of values to add is not known in advance, make a good guesstimate using whatever means, and then resize your hashtable once the number of elements added to it reaches 70% of capacity.

我希望这对你有所帮助。

PS - In C# the GetHashCode() method is pretty slow and results in actual value collisions under a lot of conditions I've tested. For some real fun, build your own hashfunction and try to get it to NEVER collide on the specific data you are hashing, run faster than GetHashCode, and have a fairly even distribution. I've done this using long instead of int size hashcode values and it's worked quite well on up to 32 million entires hashvalues in the hashtable with 0 collisions. Unfortunately I can't share the code as it belongs to my employer... but I can reveal it is possible for certain data domains. When you can achieve this, the hashtable is VERY fast. :)

2010-05-15 01:41:55