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。

你取一堆东西，和一个数组。

对于每一个东西，你为它建立一个索引，称为哈希。关于哈希的重要事情是它“分散”了很多;你不希望两个相似的东西有相似的哈希值。

你把东西放到数组中哈希值表示的位置。在一个给定的哈希中可以有多个对象，所以你可以将这些对象存储在数组或其他合适的东西中，我们通常称之为bucket。

当你在哈希中查找东西时，你会经历相同的步骤，计算哈希值，然后查看那个位置的bucket中有什么，并检查它是否是你要寻找的东西。

当你的哈希工作得很好并且你的数组足够大时，在数组的任何特定下标处最多只会有很少的东西，所以你不需要看太多。

额外的好处是，当你的哈希表被访问时，它会把找到的东西(如果有的话)移动到桶的开头，这样下次它就会是第一个被检查的东西。

哈希的计算方式通常不取决于哈希表，而是取决于添加到哈希表中的项。在框架/基类库(如。net和Java)中，每个对象都有一个GetHashCode()(或类似)方法，返回该对象的哈希码。理想的哈希码算法和准确的实现取决于对象中表示的数据。

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. :)

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