insert.h

/**
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] in as [1,5],[6,9].

Example 2:
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]. 
*/
/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<Interval> res = intervals;
        int n = res.size();
        for(int i = 0; i < n;){
            if(res[i].start > newInterval.end){
               res.insert(res.begin()+i, 1, newInterval);
               return res;
            }
            if(newInterval.start > res[i].end){
                i++;
                continue;
            }
            
            newInterval.start = min(newInterval.start, res[i].start);
            newInterval.end = max(newInterval.end, res[i].end);
            
            res.erase(res.begin()+i);
            n--;
        }
        res.push_back(newInterval);
        return res;
    }
};