算法描述：
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:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]]
Example 2:
Input: intervals =[[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval =[4,8]Output: [[1,2],[3,10],[12,16]] Explanation: Because the new interval[4,8]overlaps with[3,5],[6,7],[8,10].
解题思路：细节实现题。
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) { vector<Interval> results; int index =0; while( index < intervals.size() &&intervals[index].end < newInterval.start){ results.push_back(intervals[index++]); } while(index< intervals.size()&& intervals[index].start <= newInterval.end){ newInterval.start = min(newInterval.start,intervals[index].start); newInterval.end = max(newInterval.end,intervals[index].end); index++; } results.push_back(newInterval); while(index< intervals.size()){ results.push_back(intervals[index++]); } return results; }
原文：https://www.cnblogs.com/nobodywang/p/10370116.html