运行时间双重for循环的复杂性 [英] Running time complexity of double for-loops

问题描述

我有些由以下算法混淆。特别是，我不明白为什么第一个是O（n），第二个是O（n ^ 2）。我的唯一的直觉也许是所述内环和外环的第一算法不是链。其次，我可以直观地看出，第二种算法是O（n ^ 2），但我们将如何去寻找一些常量C1，C2证明F（n）是大0小0的n ^ 2 <？ / P>

 总和= 0;
的for（int i = 0;我n种;我++）
    为（诠释J = 0; J＆n种; J ++）
    综上所述++;
 

---

 总和= 0;
的for（int i = 0;我n种;我++）
    对于（INT J = 0; J＆LT;我; J ++）
        综上所述++;
 

解决方案

。这都是为O（n ² ）

您code有一个便利的机制，用于测量的时间复杂度，那就是之变量

去实现与不同的值 N 。如果你的之 S补充一条线，它是线性的。如果他们不它不是线性的。我想，你会发现，在你第一种算法，之将永远是完全相同 N ^ 2

I'm somewhat confused by the following algorithms. In particular, I don't understand why the first is O(n) and the second is O(n^2). My only intuition is perhaps that the inner and outer loops for the first algorithm aren't "linked." Secondly, I can intuitively see that the second algorithm is O(n^2), but how would we go about finding some constants c1, c2 to prove f(n) is big-0 and little-0 of n^2?

sum = 0;
for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
    sum++;

---

sum = 0;
for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
        sum++;

解决方案

Both of those are O(n²)

Your code has a handy mechanism for measuring the time complexity, and that is the sum variable

Go implement that with different values for n. if your sums make a line, it's linear. if they don't it's not linear. I think that you'll find that in your first algorithm, sum will always be exactly n^2

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