score:10

This can be achieved using `Enumerable.Range`

``````List<int> integerList = Enumerable.Range(MinValue, MaxValue - MinValue).ToList();
``````

score:3

Generates a sequence of integral numbers within a specified range.

A simple example, using your `MinValue` and `MaxValue` variables:

``````List<int> integerList = Enumerable.Range(MinValue, MaxValue - MinValue).ToList();
``````

Note that if `MaxValue` is less than `MinValue`, the `count` parameter will be less than zero and an `ArgumentOutOfRangeException` will be thrown.

score:1

You could use a collection initializer.

http://msdn.microsoft.com/en-us/library/bb384062.aspx

private List integerList = new List{MinValue, MaxValue};

score:2

Assuming MaxValue is always > MinValue,

``````var integerList = Enumerable.Range(MinValue, MaxValue-MinValue).ToList();
``````

score:1

You could use `Enumerable.Range`

``````List<int> integerList = Enumerable.Range(MinValue, MaxValue - MinValue).ToList();
``````

score:1

Try IEnumerable.Range(MinValue, MaxValue-MinValue).ToList(): MSDN