我是这样向我那些不懂技术的朋友描述的:

考虑多位数加法。很好的老式铅笔和纸的补充。就是你7-8岁时学的那种。给定两个三位数或四位数，你很容易就能求出它们加起来是多少。

如果我给你两个100位的数字，然后问你它们加起来是多少，即使你必须使用铅笔和纸，计算出来也会非常简单。一个聪明的孩子可以在几分钟内做这样的加法。这只需要大约100次操作。

现在，考虑多位数乘法。你可能在八九岁的时候就学会了。你(希望)做了很多重复的练习来学习它背后的机制。

Now, imagine I gave you those same two 100-digit numbers and told you to multiply them together. This would be a much, much harder task, something that would take you hours to do - and that you'd be unlikely to do without mistakes. The reason for this is that (this version of) multiplication is O(n^2); each digit in the bottom number has to be multiplied by each digit in the top number, leaving a total of about n^2 operations. In the case of the 100-digit numbers, that's 10,000 multiplications.

这可能太数学化了，但这是我的尝试。(我是数学家。)

如果某个东西是O(f(n))，那么它在n个元素上的运行时间将等于A f(n) + B(以时钟周期或CPU操作为单位)。理解这些常量A和B是非常关键的，它们来自特定的实现。B本质上代表你的操作的“常量开销”，例如你所做的一些预处理不依赖于集合的大小。A表示实际项目处理算法的速度。

关键在于，你可以使用大O符号来计算某物的可伸缩性。所以这些常数并不重要:如果你想弄清楚如何从10个项目扩展到10000个项目，谁会关心开销常数B呢?类似地，其他问题(见下文)肯定会超过乘法常数A的重要性。

So the real deal is f(n). If f grows not at all with n, e.g. f(n) = 1, then you'll scale fantastically---your running time will always just be A + B. If f grows linearly with n, i.e. f(n) = n, your running time will scale pretty much as best as can be expected---if your users are waiting 10 ns for 10 elements, they'll wait 10000 ns for 10000 elements (ignoring the additive constant). But if it grows faster, like n2, then you're in trouble; things will start slowing down way too much when you get larger collections. f(n) = n log(n) is a good compromise, usually: your operation can't be so simple as to give linear scaling, but you've managed to cut things down such that it'll scale much better than f(n) = n2.

实际上，这里有一些很好的例子:

O(1): retrieving an element from an array. We know exactly where it is in memory, so we just go get it. It doesn't matter if the collection has 10 items or 10000; it's still at index (say) 3, so we just jump to location 3 in memory. O(n): retrieving an element from a linked list. Here, A = 0.5, because on average you''ll have to go through 1/2 of the linked list before you find the element you're looking for. O(n2): various "dumb" sorting algorithms. Because generally their strategy involves, for each element (n), you look at all the other elements (so times another n, giving n2), then position yourself in the right place. O(n log(n)): various "smart" sorting algorithms. It turns out that you only need to look at, say, 10 elements in a 1010-element collection to intelligently sort yourself relative to everyone else in the collection. Because everyone else is also going to look at 10 elements, and the emergent behavior is orchestrated just right so that this is enough to produce a sorted list. O(n!): an algorithm that "tries everything," since there are (proportional to) n! possible combinations of n elements that might solve a given problem. So it just loops through all such combinations, tries them, then stops whenever it succeeds.

Big-O背后的“直觉

想象一下，当x趋于无穷时，x上的两个函数f(x)和g(x)之间的“竞争”。

现在，如果从某一点开始(某个x点)，一个函数的值总是比另一个高，那么我们称这个函数比另一个“快”。

例如，对于每x > 100，你看到f(x) > g(x)，那么f(x)比g(x)“快”。

在这种情况下，我们可以说g(x) = O(f(x))F (x)对g(x)提出了某种“速度限制”，因为最终它超过了它，并将其永远甩在后面。

这并不完全是大o符号的定义，它还指出，对于某个常数C, f(x)只需要大于C*g(x)(这只是另一种说法，你不能通过将g(x)乘以常数因子来帮助g(x)赢得竞争- f(x)最终总是会赢)。正式的定义也使用绝对值。但我希望我能让它更直观。

我试图用c#和JavaScript给出简单的代码示例来解释。

C#

For List<int> numbers = new List<int> {1,2,3,4,5,6,7,12,543,7};

O(1)看起来像

return numbers.First();

O(n)看起来像

int result = 0;
foreach (int num in numbers)
{
  result += num;
}
return result;

O(nlog (n))是这样的

int result = 0;
foreach (int num in numbers)
{
    int index = numbers.Count - 1;
    while (index > 1)
    {
        // yeah, stupid, but couldn't come up with something more useful :-(
        result += numbers[index];
        index /= 2;
    }
}
return result;

O(n2)是这样的

int result = 0;
foreach (int outerNum in numbers)
{
    foreach (int innerNum in numbers)
    {
        result += outerNum * innerNum;
    }
}
return result;

O(n!)看起来，嗯，太累了，想不出任何简单的东西。 但我希望你能明白大意?

JavaScript

对于const数= [1,2,3,4,5,6,7,12,543,7];

O(1)看起来像

numbers[0];

O(n)看起来像

let result = 0;
for (num of numbers){
    result += num;
}

O(nlog (n))是这样的

let result = 0;
for (num of numbers){

    let index = numbers.length - 1;
    while (index > 1){
        // yeah, stupid, but couldn't come up with something more useful :-(
        result += numbers[index];
        index = Math.floor(index/2)
    }
}

O(n2)是这样的

let result = 0;
for (outerNum of numbers){
    for (innerNum of numbers){
        result += outerNum * innerNum;
    }
}

大多数Jon Bentley的书(例如Programming Pearls)都以一种非常实用的方式涵盖了这些内容。他的这次演讲中就包括了一个这样的快排分析。

虽然与这个问题并不完全相关，但Knuth提出了一个有趣的想法:在高中微积分课上教授Big-O符号，尽管我觉得这个想法相当古怪。

其中很多都很容易用非编程的东西来演示，比如洗牌。

对一副牌进行排序通过遍历整副牌找到黑桃a，然后遍历整副牌找到黑桃2，以此类推最坏情况是n^2，如果这副牌已经倒着排序了。你看了52张牌52次。

一般来说，真正糟糕的算法不一定是故意的，它们通常是对其他东西的误用，比如在同一集合上线性重复的另一个方法中调用一个线性方法。