给已经好的答案锦上添花。一个实际的例子是postgres数据库中的哈希索引和b树索引。

哈希索引形成一个哈希表索引来访问磁盘上的数据，而btree顾名思义使用的是btree数据结构。

大O时间是O(1) vs O(logN)

目前不鼓励在postgres中使用哈希索引，因为在现实生活中，特别是在数据库系统中，实现无冲突的哈希是非常困难的(可能导致O(N)最坏情况的复杂性)，正因为如此，使它们具有崩溃安全性就更加困难了(在postgres中称为提前写日志- WAL)。

在这种情况下进行这种权衡，因为O(logN)对于索引来说已经足够好了，而实现O(1)非常困难，而且时间差并不重要。

总有一个隐藏常数，在O(log n)算法中可以更低。因此，在实际生活数据中，它可以更快地工作。

还有空间问题(比如在烤面包机上运行)。

还有开发人员的时间问题——O(log n)可能更容易实现和验证1000倍。

当n很小时，O(1)总是很慢。

选择大O复杂度高的算法而不是大O复杂度低的算法的原因有很多:

most of the time, lower big-O complexity is harder to achieve and requires skilled implementation, a lot of knowledge and a lot of testing. big-O hides the details about a constant: algorithm that performs in 10^5 is better from big-O point of view than 1/10^5 * log(n) (O(1) vs O(log(n)), but for most reasonable n the first one will perform better. For example the best complexity for matrix multiplication is O(n^2.373) but the constant is so high that no (to my knowledge) computational libraries use it. big-O makes sense when you calculate over something big. If you need to sort array of three numbers, it matters really little whether you use O(n*log(n)) or O(n^2) algorithm. sometimes the advantage of the lowercase time complexity can be really negligible. For example there is a data structure tango tree which gives a O(log log N) time complexity to find an item, but there is also a binary tree which finds the same in O(log n). Even for huge numbers of n = 10^20 the difference is negligible. time complexity is not everything. Imagine an algorithm that runs in O(n^2) and requires O(n^2) memory. It might be preferable over O(n^3) time and O(1) space when the n is not really big. The problem is that you can wait for a long time, but highly doubt you can find a RAM big enough to use it with your algorithm parallelization is a good feature in our distributed world. There are algorithms that are easily parallelizable, and there are some that do not parallelize at all. Sometimes it makes sense to run an algorithm on 1000 commodity machines with a higher complexity than using one machine with a slightly better complexity. in some places (security) a complexity can be a requirement. No one wants to have a hash algorithm that can hash blazingly fast (because then other people can bruteforce you way faster) although this is not related to switch of complexity, but some of the security functions should be written in a manner to prevent timing attack. They mostly stay in the same complexity class, but are modified in a way that it always takes worse case to do something. One example is comparing that strings are equal. In most applications it makes sense to break fast if the first bytes are different, but in security you will still wait for the very end to tell the bad news. somebody patented the lower-complexity algorithm and it is more economical for a company to use higher complexity than to pay money. some algorithms adapt well to particular situations. Insertion sort, for example, has an average time-complexity of O(n^2), worse than quicksort or mergesort, but as an online algorithm it can efficiently sort a list of values as they are received (as user input) where most other algorithms can only efficiently operate on a complete list of values.

并行执行算法的可能性。

我不知道是否有O(log n)和O(1)类的例子，但对于某些问题，当算法更容易并行执行时，您会选择具有更高复杂度类的算法。

有些算法不能并行化，但复杂度很低。考虑另一种算法，它可以达到相同的结果，并且可以很容易地并行化，但具有更高的复杂度类。当在一台机器上执行时，第二种算法速度较慢，但当在多台机器上执行时，实际执行时间越来越短，而第一种算法无法加快速度。

假设您正在嵌入式系统上实现一个黑名单，其中0到1,000,000之间的数字可能被列入黑名单。这就给你留下了两个选择:

使用1,000,000位的bitset 使用黑名单整数的排序数组，并使用二进制搜索来访问它们

对bitset的访问将保证常量访问。从时间复杂度来看，它是最优的。从理论和实践的角度来看(它是O(1)，常量开销极低)。

不过，你可能更喜欢第二种解决方案。特别是如果您希望黑名单整数的数量非常小，因为这样内存效率更高。

即使您不为内存稀缺的嵌入式系统开发，我也可以将任意限制从1,000,000增加到1,000,000,000,000，并提出相同的论点。那么bitset将需要大约125G的内存。保证最坏情况复杂度为O(1)可能无法说服您的老板为您提供如此强大的服务器。

在这里，我强烈倾向于二叉搜索(O(log n))或二叉树(O(log n))而不是O(1)位集。在实践中，最坏情况复杂度为O(n)的哈希表可能会击败所有这些算法。