什么是比较两个数组是否相等的最快方法？ [英] What's the fastest way to compare two arrays for equality?

问题描述

我有可能具有相同的值的对象的两个数组，但在不同的顺序，例如

I have two arrays of objects which are likely to have the same values, but in a different order, e.g.

{ "cat", "dog", "mouse", "pangolin" }

{ "dog", "pangolin", "cat", "mouse" }

我想将这两个数组相等。什么是测试这个最快的方法是什么？

I wish to treat these two arrays as equal. What's the fastest way to test this?

推荐答案

我不能保证，这是的最快的，但它肯定是非常有效的：

I can't guarantee that this is the fastest, but it's certainly quite efficient:

bool areEquivalent = array1.Length == array2.Length 
                     && new HashSet<string>(array1).SetEquals(array2);

编辑： SaeedAlg和Sandris提高对重复造成的问题，这种方法的不同频率的有效分。我可以看到两种解决方法，如果这是很重要的（没有给予过多考虑到它们各自的效率）：

SaeedAlg and Sandris raise valid points about different frequencies of duplicates causing problems with this approach. I can see two workarounds if this is important (haven't given much thought to their respective efficiencies):

1.Sort阵列，然后依次进行比较。这种方法，在理论上，应该有二次在最坏的情况下的复杂性。    例如：

1.Sort the arrays and then compare them sequentially. This approach, in theory, should have quadratic complexity in the worst case. E.g.:

return array1.Length == array2.Length
       && array1.OrderBy(s => s).SequenceEqual(array2.OrderBy(s => s));

2.Build了每个数组中字符串的频率表，然后对它们进行比较。例如：

2.Build up a frequency-table of strings in each array and then compare them. E.g.:

if(array1.Length != array2.Length)
   return false;

var f1 = array1.GroupBy(s => s)
               .Select(group => new {group.Key, Count = group.Count() });

var f2 = array2.GroupBy(s => s)
               .Select(group => new {group.Key, Count = group.Count() });

return !f1.Except(f2).Any();

这篇关于什么是比较两个数组是否相等的最快方法？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！