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;
    }