如何返回不包含在任何其他范围 C# 列表中的所有范围

Question

我有一个价值列表Tuple 如下：

List<(int, int)> values = new List<(int, int)>
{
  (12, 15),
  (18, 30),
  (18, 27),
  (27, 30)
};

现在我想删除这个列表中的对，它在更宽对的范围内。

例如：(18, 27) 和(27, 30) 在(18, 30) 的范围内。所以，我想删除它下面的那些对。

所以最终的输出一定是：

(12, 15)
(18, 30)

我正在考虑使用 LINQ 的 .RemoveAll() 方法删除这些对，但我不知道该怎么做。

任何帮助将不胜感激。

Answer 1

如果我理解正确，这将起到作用。它将列表的每个值与整个列表进行比较，直到找到另一个位于不完全相等范围内的元组。

List<(int, int)> values = new List<(int, int)>
{
    (12, 15),
    (18, 30),
    (18, 27),
    (27, 30)
};

values.RemoveAll(x => values.Any(y => x.CompareTo(y) != 0 && x.Item1 >= y.Item1 && x.Item2 <= y.Item2));

Answer 2

我相信这应该可行并且应该是 n*log(n)：

//You will see some Math.Max and Math.Min - these are required because in your comments 
//you state the pairs can be out of order (the second value is less than the first)

int upper = int.MinValue;
values = values
    //Group all pairs by their minimum value, so (18,30) and (18,27) for example 
    //would be processed together
    .GroupBy(x => Math.Min(x.Item1, x.Item2))
    //Sort the groups based on their minimum value, so they can be processed in order. 
    //Although your example does not require this, the comments say this is required.
    .OrderBy(x => x.Key)
    //From the group [which may contain (18,30) and (18,27) for example],
    //only choose the one with the highest max value. This will choose (18,30)
    .Select(x => x.OrderBy(y => Math.Max(y.Item1, y.Item2)).Last())
    //Now iterate through the items and throw away the items which are inside
    //a pair previously processed pair. We use the upper variable to
    //keep track of the greatest processed upper bound to make throwing away 
    //previously processed pairs easy
    .Where(x =>
    {
        var max = Math.Max(x.Item1, x.Item2);
        if (upper < max)
        {
            upper = max;
            return true;
        }
        else
        {
            return false;
        }

    }).ToList();

如果没有 Linq，这当然可以更快，这取决于内存和性能对您的应用程序的重要性。