简短而甜蜜:

哈希表封装了一个数组，我们称之为internalArray。将项以如下方式插入数组:

let insert key value =
    internalArray[hash(key) % internalArray.Length] <- (key, value)
    //oversimplified for educational purposes

有时两个键会散列到数组中的同一个索引，而您希望保留这两个值。我喜欢把两个值都存储在同一个索引中，通过将internalArray作为一个链表数组来编码很简单:

let insert key value =
    internalArray[hash(key) % internalArray.Length].AddLast(key, value)

所以，如果我想从哈希表中检索一个项，我可以这样写:

let get key =
    let linkedList = internalArray[hash(key) % internalArray.Length]
    for (testKey, value) in linkedList
        if (testKey = key) then return value
    return null

删除操作写起来也很简单。正如你所知道的，从我们的链表数组中插入、查找和删除几乎是O(1)。

当我们的internalArray太满时，可能在85%左右的容量，我们可以调整内部数组的大小，并将所有项目从旧数组移动到新数组中。

这是一个相当深奥的理论领域，但基本轮廓很简单。

本质上，哈希函数只是一个函数，它从一个空间(比如任意长度的字符串)获取内容，并将它们映射到一个用于索引的空间(比如无符号整数)。

如果你只有一个小空间的东西来散列，你可能只需要把这些东西解释为整数，你就完成了(例如4字节字符串)

不过，通常情况下，你的空间要大得多。如果你允许作为键的空间大于你用于索引的空间(你的uint32或其他)，那么你不可能为每个键都有唯一的值。当两个或多个东西散列到相同的结果时，您必须以适当的方式处理冗余(这通常被称为冲突，如何处理它或不处理它将略微取决于您使用散列的目的)。

这意味着你不希望得到相同的结果，你也可能希望哈希函数是快速的。

平衡这两个属性(以及其他一些属性)让许多人忙得不可开交!

在实践中，您通常应该能够找到一个已知适合您的应用程序的函数并使用它。

Now to make this work as a hashtable: Imagine you didn't care about memory usage. Then you can create an array as long as your indexing set (all uint32's, for example). As you add something to the table, you hash it's key and look at the array at that index. If there is nothing there, you put your value there. If there is already something there, you add this new entry to a list of things at that address, along with enough information (your original key, or something clever) to find which entry actually belongs to which key.

因此，随着时间的推移，哈希表(数组)中的每个条目要么是空的，要么包含一个条目，要么包含一个条目列表。检索很简单，就像在数组中建立索引，然后返回值，或者遍历值列表并返回正确的值。

当然，在实践中你通常不能这样做，它浪费太多的内存。因此，所有操作都基于稀疏数组(其中唯一的条目是实际使用的条目，其他所有内容都隐式为空)。

有很多方案和技巧可以让它更好地工作，但这是最基本的。

到目前为止，所有的答案都很好，并且从不同的方面了解了哈希表的工作方式。这里有一个简单的例子，可能会有帮助。假设我们想要存储一些带有小写字母字符串的项作为键。

正如simon所解释的，哈希函数用于从大空间映射到小空间。对于我们的例子，一个简单的哈希函数实现可以取字符串的第一个字母，并将其映射为一个整数，因此“短吻鳄”的哈希代码为0，“蜜蜂”的哈希代码为1，“斑马”的哈希代码为25，等等。

接下来，我们有一个包含26个存储桶的数组(在Java中可以是数组列表)，我们将项放入与键的哈希码匹配的存储桶中。如果我们有不止一个元素键以相同字母开头，它们就会有相同的哈希码，所以它们都会进入存储桶中寻找那个哈希码所以必须在存储桶中进行线性搜索才能找到一个特定的元素。

在我们的例子中，如果我们只有几十个项目，键横跨字母表，它会工作得很好。然而，如果我们有一百万个条目，或者所有的键都以'a'或'b'开头，那么我们的哈希表就不是理想的。为了获得更好的性能，我们需要一个不同的哈希函数和/或更多的桶。

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. :)

简短而甜蜜:

哈希表封装了一个数组，我们称之为internalArray。将项以如下方式插入数组:

let insert key value =
    internalArray[hash(key) % internalArray.Length] <- (key, value)
    //oversimplified for educational purposes

有时两个键会散列到数组中的同一个索引，而您希望保留这两个值。我喜欢把两个值都存储在同一个索引中，通过将internalArray作为一个链表数组来编码很简单:

let insert key value =
    internalArray[hash(key) % internalArray.Length].AddLast(key, value)

所以，如果我想从哈希表中检索一个项，我可以这样写:

let get key =
    let linkedList = internalArray[hash(key) % internalArray.Length]
    for (testKey, value) in linkedList
        if (testKey = key) then return value
    return null

删除操作写起来也很简单。正如你所知道的，从我们的链表数组中插入、查找和删除几乎是O(1)。

当我们的internalArray太满时，可能在85%左右的容量，我们可以调整内部数组的大小，并将所有项目从旧数组移动到新数组中。

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.