A more general question is if there are situations where one would prefer an O(f(n)) algorithm to an O(g(n)) algorithm even though g(n) << f(n) as n tends to infinity. As others have already mentioned, the answer is clearly "yes" in the case where f(n) = log(n) and g(n) = 1. It is sometimes yes even in the case that f(n) is polynomial but g(n) is exponential. A famous and important example is that of the Simplex Algorithm for solving linear programming problems. In the 1970s it was shown to be O(2^n). Thus, its worse-case behavior is infeasible. But -- its average case behavior is extremely good, even for practical problems with tens of thousands of variables and constraints. In the 1980s, polynomial time algorithms (such a Karmarkar's interior-point algorithm) for linear programming were discovered, but 30 years later the simplex algorithm still seems to be the algorithm of choice (except for certain very large problems). This is for the obvious reason that average-case behavior is often more important than worse-case behavior, but also for a more subtle reason that the simplex algorithm is in some sense more informative (e.g. sensitivity information is easier to extract).

以下是我的观点:

有时，当算法在特定的硬件环境中运行时，会选择较差的复杂度算法来代替较好的算法。假设我们的O(1)算法非顺序地访问一个非常大的固定大小数组的每个元素来解决我们的问题。然后将该阵列放在机械硬盘驱动器或磁带上。

在这种情况下，O(logn)算法(假设它按顺序访问磁盘)变得更有利。

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。

考虑一个红黑树。它具有O(log n)的访问、搜索、插入和删除操作。与数组相比，数组的访问权限为O(1)，其余操作为O(n)。

因此，对于一个插入、删除或搜索比访问更频繁的应用程序，并且只能在这两种结构之间进行选择，我们更喜欢红黑树。在这种情况下，你可能会说我们更喜欢红黑树更麻烦的O(log n)访问时间。

为什么?因为权限不是我们最关心的。我们正在权衡:应用程序的性能更大程度上受到其他因素的影响。我们允许这种特定的算法受到性能影响，因为我们通过优化其他算法获得了很大的收益。

So the answer to your question is simply this: when the algorithm's growth rate isn't what we want to optimize, when we want to optimize something else. All of the other answers are special cases of this. Sometimes we optimize the run time of other operations. Sometimes we optimize for memory. Sometimes we optimize for security. Sometimes we optimize maintainability. Sometimes we optimize for development time. Even the overriding constant being low enough to matter is optimizing for run time when you know the growth rate of the algorithm isn't the greatest impact on run time. (If your data set was outside this range, you would optimize for the growth rate of the algorithm because it would eventually dominate the constant.) Everything has a cost, and in many cases, we trade the cost of a higher growth rate for the algorithm to optimize something else.

简单地说:因为系数(与该步骤的设置、存储和执行时间相关的成本)在较小的大o问题中比在较大的大o问题中要大得多。Big-O只是算法可伸缩性的一个衡量标准。

考虑以下来自黑客词典的例子，提出了一个依赖于量子力学的多重世界解释的排序算法:

用量子过程随机排列数组， 如果数组没有排序，毁灭宇宙。 所有剩下的宇宙现在都被排序了(包括你所在的宇宙)。

(来源:http://catb.org/ esr /术语/ html / B / bogo-sort.html)

注意，这个算法的大O是O(n)，它击败了迄今为止在一般项目上的任何已知排序算法。线性阶跃的系数也很低(因为它只是一个比较，而不是交换，是线性完成的)。事实上，类似的算法可以用于在多项式时间内解决NP和co-NP中的任何问题，因为每个可能的解(或没有解的可能证明)都可以使用量子过程生成，然后在多项式时间内验证。

然而，在大多数情况下，我们可能不想冒多重世界可能不正确的风险，更不用说实现步骤2的行为仍然是“留给读者的练习”。

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.