Merge Intervals

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 [1,3],[6,9] insert and merge [2,5] would result in [1,5],[6,9].

Example 2:

Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] would result in [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

Make sure the returned intervals are also sorted.


*******************************************************************************************************
/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
  static bool myfunc(const Interval &a, const Interval &b){
        return (a.start < b.start);
    }
    
vector<Interval> Solution::insert(vector<Interval> &intervals, Interval newInterval) {
    intervals.push_back(newInterval);
    vector<Interval> res;
        if (intervals.size()==0){return res;}
        sort(intervals.begin(),intervals.end(),myfunc);
        res.push_back(intervals[0]);
        for (int i=1;i<intervals.size();i++){
            if (res.back().end>=intervals[i].start){
                res.back().end=max(res.back().end,intervals[i].end);
            }else{
                res.push_back(intervals[i]);
            }   
        }
        return res;
    }