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n 元搜索的时间复杂度。

[英]Time complexity for n-ary search.

原文 2015-11-16 02:58:38 1 1 algorithm/ time-complexity/ binary-search

I am studying time complexity for binary search, ternary search and k-ary search in N elements and have come up with its respective asymptotic worse case run- time.我正在研究 N 元素中二元搜索、三元搜索和 k 元搜索的时间复杂度，并提出了其各自的渐近最坏情况运行时间。 However, I started to wonder what would happen if I divide N elements into n ranges (or aka n-ary search in n elements).但是，我开始想知道如果我将 N 个元素分成 n 个范围（或 n 个元素中的 n 元搜索）会发生什么。 Would that be a sorted linear search in an array which would result in a run-time of O(N)?这会是一个数组中的排序线性搜索，它会导致 O(N) 的运行时间吗？ This is a little confusing.这有点令人困惑。 Please help me out!请帮帮我！

1 个解决方案

What you say is right.你说的是对的。

For a k-ary search we have:对于k-ary搜索，我们有：

Do k-1 checks in boundaries to isolate one of the k ranges.在边界中执行k-1检查以隔离k范围之一。
Jump into the range obtained from above.跳入从上面获得的范围。

Hence the time complexity is essentially O((k-1)*log_k(N)) where log_k(N) means ' log(N) to base k '.因此，时间复杂度本质上是O((k-1)*log_k(N)) ，其中log_k(N)表示' log(N)到基数k '。 This has a minimum when k=2 .当k=2时，这具有最小值。

If k = N , the time complexity will be: O((N-1) * log_N(N)) = O(N-1) = O(N) , which is the same algorithmically and complexity-wise as linear search.如果k = N ，时间复杂度将为： O((N-1) * log_N(N)) = O(N-1) = O(N) ，这在算法和复杂度方面与线性搜索相同。

Translated to the algorithm above, it is:转化为上面的算法，就是：

Do N-1 checks in boundaries (each of the first N-1 elements) to isolate one of the N ranges.在边界（前N-1元素中的每一个）中执行N-1检查以隔离N范围之一。 This is the same as a linear search in the first N-1 elements.这与在前N-1元素中的线性搜索相同。
Jump into the range obtained from above.跳入从上面获得的范围。 This is the same as checking the last element (in constant time).这与检查最后一个元素（在恒定时间内）相同。