log(n) means logarithmic growth. An example would be divide and conquer algorithms. If you have 1000 sorted numbers in an array ( ex. 3, 10, 34, 244, 1203 ... ) and want to search for a number in the list (find its position), you could start with checking the value of the number at index 500. If it is lower than what you seek, jump to 750. If it is higher than what you seek, jump to 250. Then you repeat the process until you find your value (and key). Every time we jump half the search space, we can cull away testing many other values since we know the number 3004 can't be above number 5000 (remember, it is a sorted list).

N log(N)表示N * log(N)

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。

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;
    }
}

不，O(n)算法并不意味着它将对每个元素执行操作。大o符号给了你一种方法来谈论你的算法的“速度”独立于你的实际机器。

O(n)表示算法花费的时间随着输入的增加而线性增长。O(n²)意味着你的算法花费的时间是你输入的平方。等等。

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。

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.