1 回答
Your question and examples imply non-overlapping intervals. In this case you can just perform a binary search - whether comparison is done by start time or end time does not matter for non-overlapping intervals - and insert the new interval at the position found if not already present.
UPDATE
I missed the merging occurring in the first example. A bad case is inserting a large interval into a long list of short intervals where the long interval overlaps many short intervals. To avoid a linear search for all intervals that have to be merged one could perform two binary searches - one from the left comparing by start time and one from the right comparing by the end time.
Now it is trivial to decide if the interval is present, must be inserted or must be merged with the intervals between the positions found by the two searches. While this is not very complex it is probably very prone to off-by-one errors and requires some testing.