我经常用O(1/n)来描述随着输入变大而变小的概率——例如，在log2(n)次投掷中，一枚均匀硬币背面朝上的概率是O(1/n)。

你不能低于O(1)但是O(k) k小于N是可能的。我们称之为次线性时间算法。在某些问题中，次线性时间算法只能给出特定问题的近似解。然而，有时，一个近似解就可以了，可能是因为数据集太大了，或者计算所有数据的计算成本太高了。

O(1)仅仅表示“常数时间”。

当你给循环[1]添加一个早期退出时，你(在大O符号中)把一个O(1)算法变成了O(n)算法，但使它更快。

诀窍是一般情况下，常数时间算法是最好的，线性算法比指数算法好，但对于n很小的时候，指数算法可能更快。

1:假设这个例子的列表长度是静态的

不，这不可能:

随着n在1/n范围内趋于无穷，我们最终得到1/(无穷)，这实际上是0。

因此，问题的大-oh类将是O(0)和一个巨大的n，但更接近常数时间和一个低n。这是不明智的，因为唯一可以在比常数时间更快的时间内完成的事情是:

Void nothing() {};

甚至这也是有争议的!

只要你执行了一个命令，你至少在O(1)，所以不，我们不能有一个O(1/n)的大哦类!

It may be possible to construct an algorithm that is O(1/n). One example would be a loop that iterates some multiple of f(n)-n times where f(n) is some function whose value is guaranteed to be greater than n and the limit of f(n)-n as n approaches infinity is zero. The calculation of f(n) would also need to be constant for all n. I do not know off hand what f(n) would look like or what application such an algorithm would have, in my opinion however such a function could exist but the resulting algorithm would have no purpose other than to prove the possibility of an algorithm with O(1/n).

其余的大多数答案都将大o解释为专门关于算法的运行时间。但是因为问题没有提到它，我认为值得一提的是大o在数值分析中的另一个应用，关于误差。

Many algorithms can be O(h^p) or O(n^{-p}) depending on whether you're talking about step-size (h) or number of divisions (n). For example, in Euler's method, you look for an estimate of y(h) given that you know y(0) and dy/dx (the derivative of y). Your estimate of y(h) is more accurate the closer h is to 0. So in order to find y(x) for some arbitrary x, one takes the interval 0 to x, splits it up until n pieces, and runs Euler's method at each point, to get from y(0) to y(x/n) to y(2x/n), and so on.

欧拉方法是O(h)或O(1/n)算法，其中h通常被解释为步长n被解释为你划分一个区间的次数。

在实际数值分析应用中，由于浮点舍入误差，也可以有O(1/h)。你的间隔越小，某些算法的实现就会抵消得越多，丢失的有效数字就越多，因此在算法中传播的错误也就越多。

For Euler's method, if you are using floating points, use a small enough step and cancellation and you're adding a small number to a big number, leaving the big number unchanged. For algorithms that calculate the derivative through subtracting from each other two numbers from a function evaluated at two very close positions, approximating y'(x) with (y(x+h) - y(x) / h), in smooth functions y(x+h) gets close to y(x) resulting in large cancellation and an estimate for the derivative with fewer significant figures. This will in turn propagate to whatever algorithm you require the derivative for (e.g., a boundary value problem).