How to get a distinct list in post order (last value)?-CodePudding

.Distinct() returns distinct values of a list in order, I.e. 1,2,3,1 becomes 1,2,3.

I want to get the distinct values of a last but always return the last value instead of the first - how would i do that?

var input = new List<string> { "AAA", "BBB", "CCC", "AAA" };

//input = input
//    .Distinct()
//    .ToList(); 
// Returns: "AAA", "BBB", "CCC"

//input = input
//    .GroupBy(c => c)
//    .Select(c => c.OrderByDescending(x => x).Last())
//    .ToList();
// Returns: "AAA", "BBB", "CCC"

// TEST
var output = new List<string> { "BBB", "CCC", "AAA" };

if (input.SequenceEqual(output))
{
    Console.WriteLine("Success!");
}
else
{
    Console.WriteLine("Fail!");
}

CodePudding user response：

As Lasse v. Karlsen pointed out, reversing the string running a distinct and reversing the string again works.

input.Reverse();
input = input
    .Distinct()
    .Reverse()
    .ToList();

Just be mindful that 2 different .Revers() calls are used: List<T>.Reverse() and Enumerable.Reverse(source) (Linq).