Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals
Given intervals
[1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given
Given
[1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
This is because the new interval
[4,9] overlaps with [3,5],[6,7],[8,10].
One consideration is, the i-place method maybe slower than using extra space for the new vector. So I create a new vector res for result returned.
Linear scan, and keep updating newInterval until intervals[i].start is larger than newInterval.end.
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