Given a set ofnon-overlappingintervals, 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.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> ans = new ArrayList<>();

        int i = 0;
        while (i < intervals.size()){
            if (intervals.get(i).end < newInterval.start){
                ans.add(intervals.get(i));
                i++;
            }
            else{
                break;
            }
        }

        while (i < intervals.size() && intervals.get(i).start <= newInterval.end){
            newInterval = new Interval(
                Math.min(intervals.get(i).start, newInterval.start), 
                Math.max(intervals.get(i).end, newInterval.end)
            );
            i++;
        }
        ans.add(newInterval);

        while (i < intervals.size()){
            ans.add(intervals.get(i));
            i++;
        }

        return ans;
    }
}