blocks|key|756729|text|他说，因为10N+%2B8是一个线性方程。如果你画出这个方程，你就会得到一条直线。试着在这个网站上输入10+*+x+%2B+8+(函数图)，自己看看。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|756730|entityMap|0|LINK|mutability|MUTABLE|url|http://rechneronline.de/function-graphs/^0|1D|A|1P|3|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

He said that because 10N + 8 is a linear equation. If you plot that equation you get a straight line. Try typing <code>10 * x + 8</code> on this website (<a href="http://rechneronline.de/function-graphs/">function graphs</a>) and see for yourself.

blocks|key|609994|text|作者只是根据自己的经验选择最合适的。您应该知道，计数算法的复杂性几乎总是意味着找到一个大欧函数，它反过来只是给定函数的上界(在您的例子中是10N%2B8)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|609995|已知的复杂性类型只有几种:线性复杂度、二次复杂度、等.因此，计算时间复杂度的最后一步是选择不太复杂的类型(我的意思是，线性比二次复杂，二次不那么复杂，指数等)可以用于给定的函数，从而正确地描述其复杂性。|609996|在您的例子中，O(n)和O(n%5E2)甚至O(2%5En)确实是的正确答案。但是，在大字号定义中最适合的不太复杂的函数是O(n)，这是这里的一个答案。|609997|这是一个真好文章，充分解释了大噢符号.|609998|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Time_complexity|1|http://en.wikipedia.org/wiki/Big_O_notation|2|http://www.csee.umbc.edu/~chang/cs202.f98/readings/asymptotic/asymptotic.html^0|17|2|1N|2|1X|5|0|P|1|0|0|T|5|1L|4|13|5|1|0|E|4|4|4|2|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@$9|15|A|16|1|17]]|E|$]]|$1|H|3|I|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|J|3|K|5|6|7|1G|8|@$9|1H|A|1I|B|C]]|D|@$9|1J|A|1K|1|1L]]|E|$]]|$1|L|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|M|$N|$5|O|P|Q|E|$R|S]]|T|$5|O|P|Q|E|$R|U]]|V|$5|O|P|Q|E|$R|W]]]]

The author just based on his experience in choosing most appropriate . You should know, that counting algorithm complexity almost always means to find a Big-Oh function, which, in turn, is just an upper bound to given function (10N+8 in your case).

There is only a few well-known complexity types: linear complexity, quadratic complexity, <a href="http://en.wikipedia.org/wiki/Time_complexity" rel="nofollow">etc</a>. So, the final step of counting a time complexity consists of choosing what is the less complex type (i mean, linear is less complex than a quadratic, and quadratic is less complex that exponential, and so on) can be used for the given function, which correctly describe its complexity.

In your case, O(n) and O(n^2) and even O(2^n) are the right answers indeed. But the less complex function, which suits perfectly in <a href="http://en.wikipedia.org/wiki/Big_O_notation" rel="nofollow">Big-Oh notation definition</a> is O(n), which is an answer here.

Here is a <a href="http://www.csee.umbc.edu/~chang/cs202.f98/readings/asymptotic/asymptotic.html" rel="nofollow">real good article</a>, fully explained Big-Oh notation.

blocks|key|756797|text|时间复杂性的上升顺序(常见的)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|756798|O(1)+-+Constant
O(log+n)+-+logarithmic
O(n)+-+linear
O(n+log+n)+-+loglinear
O(n%5E2)+-+quadratic|code-block|syntax|javascript|756799|注:n无界增加|756800|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Ascending order of time complexities(the common one)

<pre><code>O(1) - Constant
O(log n) - logarithmic
O(n) - linear
O(n log n) - loglinear
O(n^2) - quadratic
</code></pre>

Note: N increases without bounds

blocks|key|1235796|text|对于复杂性理论，你一定要阅读一些背景理论。它通常是关于渐近复杂性的，这就是为什么您可以放弃较小的部分，而只保留复杂性类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1235797|关键的想法是，N和N%2B5之间的差异一旦N变得很大，就变成了否定的。|offset|length|style|CODE|1235798|有关更多细节，请在这里阅读：|1235799|符号表示法|1235800|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Big_O_notation^0|0|7|1|9|3|J|1|0|0|0|5|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@$D|V|E|W|F|G]|$D|X|E|Y|F|G]|$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|11|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|12|8|@]|9|@$D|13|E|14|1|15]]|A|$]]|$1|L|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

For complexity theory you definitely should read some background theory. It's usually about asymptotic complexity, which is why you can drop the smaller parts, and only keep the complexity class.

The key idea is that the difference between <code>N</code> and <code>N+5</code> becomes neglibile once <code>N</code> is really big.

For more details, start reading here:

<a href="http://en.wikipedia.org/wiki/Big_O_notation" rel="nofollow">http://en.wikipedia.org/wiki/Big_O_notation</a>

blocks|key|610025|text|一个非常务实的规则是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|610026|当一个算法的复杂度被一个类似于的A*n%5E2%2BB*n%2BC+表示时，那么复杂度(即O(某某物))的阶数等于变量n的最高阶。|offset|length|style|BOLD|CODE|610027|在A*n%5E2%2BB*n%2BC聚类中，阶数为O(n%5E2)。|610028|就像josnidhin解释的，如果聚|610029|1阶(即n)-它称为线性|unordered-list-item|610030|2阶(即n%5E2)+-它称为二次型。|610031|..。诸若此类。|610032|entityMap^0|0|0|F|S|9|19|9|G|B|0|1|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@$D|X|E|Y|F|G]|$D|Z|E|10|F|G]|$D|11|E|12|F|G]|$D|13|E|14|F|H]]|9|@]|A|$]]|$1|I|3|J|5|6|7|15|8|@$D|16|E|17|F|H]]|9|@]|A|$]]|$1|K|3|L|5|6|7|18|8|@]|9|@]|A|$]]|$1|M|3|N|5|O|7|19|8|@]|9|@]|A|$]]|$1|P|3|Q|5|O|7|1A|8|@]|9|@]|A|$]]|$1|R|3|S|5|O|7|1B|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|U|$]]

A very pragmatic rule is:

when the complexity of an algorithm si represented by a poly like <code>A*n^2+B*n+C</code> then the order of complexity ( that is to say the O(something) ) is equal to the highest order of the variable n. 

In the <code>A*n^2+B*n+C</code> poly the order is O(n^2).

Like josnidhin explained, if the poly has

<ul>
<li>order 1 (i.e. n)- it is called linear</li>
<li>order 2 (i.e. n^2) - it is called quadratic</li>
<li>... and so on.</li>
</ul>

Today i'm come across with the blog in msdn and i noticed how to calculate the time complexity of an algorithm. I perfectly understand how to calculate the time complexity of an algorithm but in the last the author mentioned the below lines 

<blockquote>
 Adding everything up I get
 
 (N+4)+(5N+2)+(4N+2) = 10N+8
 
 So the asymptotic time complexity for the above code is O(N), which
 means that the above algorithm is a liner time complexity algorithm.
</blockquote>

So how come the author is said it's based on the liner time complexity algorithm. The link for the blog 

<a href="http://blogs.msdn.com/b/nmallick/archive/2010/03/30/how-to-calculate-time-complexity-for-a-given-algorithm.aspx" rel="nofollow">http://blogs.msdn.com/b/nmallick/archive/2010/03/30/how-to-calculate-time-complexity-for-a-given-algorithm.aspx</a>.

Time Complexity of an algorithm : How to decide which algorithm after calculated the time

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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今天，我在msdn中看到了博客，我注意到了如何计算算法的时间复杂度。我完全理解如何计算算法的时间复杂度，但在最后，作者提到了以下几行把我得到的一切加起来(N+4)+(5N+2)+(4N+2) = 10N+8因此，上述算法的渐近时间复杂度为O(N)，这意味着上述算法是线性时间复杂度算法。为什么说它是基于线性时间复杂度算法的呢？博客的链接。

问算法的时间复杂度:如何在计算时间后决定哪一种算法
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