这种方法可以检查数组中的重复项比较有效吗？为什么或者为什么不？ [英] Is this approach to check duplicates in an array relatively efficient? Why or why not?

问题描述

我只想检查我的数组中是否包含任何重复项。我在Google上搜索并看到一些方法：

I just wanted to check whether there contains any duplicates in my array. I searched on Google and see some approaches:

• 通过数组进行双重循环循环，并比较每个项目


• 创建一个存储每个项目出现次数的字典

但是这些方法需要很多循环，这种懒惰为这个功能写了大量的代码。 xD。

But these methods require a lot of loops and I'm kind of lazy to write a large amount of code just for this functionality. xD.

所以我想到了这种创意的方式：

So I thought of this creative way:

let containsDuplicates = Set(array).count != array.count

但是，这种方法是更快还是慢比其他两个？我不确定，因为它似乎创建一个我认为需要循环通过数组的集合。我不知道访问计数是否也循环遍历整个数组。

However, is this method faster or slower than the other two? I'm not sure because it seems to create a set which I think needs to loop through the array. And I don't know whether accessing the count also loops through the whole array.

如果我只有数组中最多的50个项目，这个甚至是重要的吗？

If I only have at most 50 items in the array, will this even matter?

推荐答案

这几乎是最好的方法。字典存储是恒定时间，每个项目只需要存储一次，因此它是一个 O（n）解决方案，而$ code> O（n ^ 2）用于双循环方法。空间效率较低，但这几天通常不是一个问题。

That's pretty much the best way. Dictionary storage is constant time, and each item only needs to be stored once, so it's an O(n) solution, vs the O(n^2) for the double for-loop approach. The space efficiency is lower, but that's usually not a concern these days.

如果你经常这样做，我会建议使用某种混合数据结构，所以你不是不断地生成这些从头开始设置。

If you're doing this often, I would suggest using some kind of hybrid data structure, so that you're not constantly generating these Sets from scratch.

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