Объединение двух списков без дублирования

Question

Мне нужно создать один список на основе двух других списков.Но похоже, что он не удаляет дубликаты.

Это эффективный способ объединения двух списков без дублирования?

List<String[]> blocksComparisonSet1 = new List<String[]>(); 
List<String[]> blocksComparisonSet2 = new List<String[]>(); 
//we will combine list1 and list2 into this one
List<String[]> blocksComparisonFinal = new List<String[]>(); 

//this is how I store data in each list  
//if both values found, add both of them (partial functions, FYI)
String[] NA = new String[2]; //keep results
NA[0] = baseLine; //[0] for base
NA[1] = resultLine; //[1] for result
blocksComparisonSet1.Add(NA);

//if only one value found
String[] NA = new String[2]; //keep results
NA[0] = ""; //[0] for base
NA[1] = resultLine; //[1] for result
blocksComparisonSet1.Add(NA);

//This is where I merge lists and try to remove duplicates
if (blocksComparisonSet1.Count() > 0 || blocksComparisonSet2.Count() > 0) 
//check if we have any values in out differences lists. if we do, merge them
{
    blocksComparisonFinal.AddRange(blocksComparisonSet1); 
    //add records from one list to final list
    blocksComparisonFinal.AddRange(blocksComparisonSet2); 
    //add records from second list to final list
    blocksComparisonFinal = blocksComparisonFinal.Distinct().ToList(); 
    //remove dublicates
}

-

List1[] Example    
string1 na
string2 na
string3 String1
string4 String7
string5 string8
na  string9
na  string2

-

List2[]  Example    
na  string2
na  string5
String1 string3
String7 string4
string8 string5
string9 na
string2 na

-

Final List[] must be:
na  string9
na  string2
na  string5
string1 na
String1 string3
string2 na
string3 String1
string4 String7
string5 string8
String7 string4
string8 string5
string9 na

Answer 1

var merge = blocksComparisonSet1.Union(
                blocksComparisonSet2,
                new ArrayEqualityComparer<string>()
            ).ToList();

Вам понадобится кастом IEqualityComparer<string[]>.См. MSDN .

Вот одна реализация:

class ArrayEqualityComparer<T> : IEqualityComparer<T[]> {
    public bool Equals(T[] x, T[] y) {
        if(Object.ReferenceEquals(x, y)) {
            return true;
        }
        if(x == null || y == null) {
            return false;
        }
        return x.SequenceEqual(y);
    }

    public int GetHashCode(T[] x) {
        if(x == null) {
            return 0; 
        }
        return x.Aggregate(
            0,
            (h, item) => h ^ (item != null ? item.GetHashCode() : 0)
        );
    }
}

Answer 2

Альтернативное решение, отличное от Linq, с реализацией IEqualityComparer<string[]>

var merged = new HashSet<string[]>(blocksComparisonSet1, new SEC());
merged.UnionWith(blocksComparisonSet2);

class SEC : IEqualityComparer<string[]>
{
    public bool Equals(string[] p1, string[] p2){
        return p1.SequenceEqual(p2);
    }
    public int GetHashCode(string[] p){ 
        return (int)p.Sum (p1 => p1.GetHashCode()); 
    }
}