问题描述
我需要检查两个数组是否包含任何顺序相同的数据。
利用虚比较的方法，我想这样做：
I need to check whether two arrays contain the same data in any order.Using the imaginary compare method, I would like to do:
arr1 = [1,2,3,5,4]
arr2 = [3,4,2,1,5]
arr3 = [3,4,2,1,5,5]
arr1.compare(arr2) #true
arr1.compare(arr3) #false
我用 arr1.sort == arr2.sort ，这似乎是工作，但有这样做的更好的办法？
I used arr1.sort == arr2.sort, which appears to work, but is there a better way of doing this?
推荐答案
比较它们之前排序数组是O（n log n）的。此外，维克多所指出的，即使你的数组包含非排序的对象遇到麻烦。它的速度更快，比较直方图，为O（n）。
Sorting the arrays prior to comparing them is O(n log n). Moreover, as Victor points out, you'll run into trouble if the array contains non-sortable objects. It's faster to compare histograms, O(n).
您会发现，但实现它自己，这是pretty简单，如果你preFER避免增加更多的依赖关系：
You'll find Enumerable#frequency in Facets, but implement it yourself, which is pretty straightforward, if you prefer to avoid adding more dependencies:
require 'facets'
[1, 2, 1].frequency == [2, 1, 1].frequency
#=> true
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