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
    
      insert(vector
     
       &ints, Interval newInterval) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector
      
        intervals( ints);                        int low = newInterval.start;        int high = newInterval.end;                bool delFlag = false;                for(int i=0; i
       
         low ) {                if( intervals[i].end >= low) {                    newInterval.start = intervals[i].start;                    intervals.erase( intervals.begin() + i );                }                 delFlag = true;                    }                        if( delFlag) {                    intervals.erase( intervals.begin() + i);            }                        if( intervals[i].end >= high ) {                if( intervals[i].start <= high) {                    newInterval.end = intervals[i].end;                    intervals.erase( intervals.begin() + i);                }                delFlag = false;                                intervals.insert( newInterval, i-1);            }        }        return intervals;    }};