据此，在哈希表中搜索的时间复杂度为 O(1)。
但是，如果发生碰撞，那么显然这应该是 O(1) + 一些东西。
我的问题是：
当你说
get(someKey)
从散列表中，散列函数应用于 someKey，并直接从该位置检索数据。
但是想象一下分离链接用于冲突解决。并想象 someKey 和 someOtherKey 在我们的散列函数应用于它们之后具有相同的输出。假设它是值“25”。
所以当我说
get(someKey)
我将从位置“25”获取数据。这使它成为 O(1)。伟大的。
然而当我说
get(someOtherKey)
现在someOtherKey链接到someKey所在的位置。
当对someOtherKey应用散列时，我得到 25。
我如何获得我需要的价值？内部是什么？还有别的表吗？算法流程如何？是否有其他表用于存储所有碰撞？
谢谢你。我希望我的问题很清楚！
有许多不同的数据结构可以用于处理冲突。这是一个很好的总结。http://en.wikipedia.org/wiki/Hash_table。
The hash function narrows the search down to a single bucket in the data structure. The bucket then contains another data structure for resolving collisions. It can be link to an array, where the keys are maintained in a sorted or unsorted order. The link might be to the first element in a linked list of keys, or to the root node of a b-tree. The important point is that the hashing function very quickly narrows down the scope of the search.
Once the scope is narrowed, some other less-efficient search can be useful to work through the collisions. It's all about the trade-offs. You want a hashing algorithm that gives a large enough range of hashes (and buckets) to minimize collisions, limited to how much memory you can afford. If the collisions are rare, a linear search through a linked list of collisions isn't bad. If there many collisions, then the efficiency of re-sizing arrays for the buckets becomes more important.