比较两个数组是否相等的最快方法是什么?

我不能保证这是 ,但它确实非常有效:

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

编辑: SaeedAlg和Sandris提出了不同重复频率的有效点,从而导致了这种方法的问题。如果这一点很重要的话,我可以看到两个解决方案(没有考虑它们各自的效率):

例如。:

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

2、建立每个数组中字符串的频率表,并进行比较。例如。:

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();

我认为唯一合理的方法是把它们分类然后比较。

O(n logn) 和比较 O(n) O(对数)

你试过类似的东西吗

string[] arr1 = {"cat", "dog", "mouse", "pangolin"};

string[] arr2 = {"dog", "pangolin", "cat", "mouse"};

bool equal = arr1.Except(arr2).Count() == 0 && arr2.Except(arr1).Count() == 0;

假设没有重复项,我将使用哈希集

string[] arr1 = new string[] { "cat", "dog", "mouse", "pangolin" };
string[] arr2 = new string[] { "dog", "pangolin", "cat", "mouse" };

bool result = true;
if (arr1.Length != arr2.Length)
{
    result = false;
}
else
{
    HashSet<string> hash1 = new HashSet<string>(arr1);
    foreach (var s in arr2)
    {
        if (!hash1.Contains(s))
            result = false;
    }
}

编辑:

伪码:

A:array
B:array
C:hashtable

if A.length != B.length then return false;

foreach objA in A
{
H = objA;
if H is not found in C.Keys then
C.add(H as key,1 as initial value);
else
C.Val[H as key]++;
}

foreach objB in B
{
H = objB;
if H is not found in C.Keys then
return false;
else
C.Val[H as key]--;
}

if(C contains non-zero value)
return false;
else
return true;