# Java solution with comments (revised based on problem 56. Merge intervals)

• I did "Merge Intervals" problem first. When I came to this problem, I realized that I only need to insert the new interval in the list of intervals, and call the function "merge" that I wrote in "Merge Intervals".
Not sure if this would be a good answer, but if you have done Merge Intervals, I guess you should be able to get this solution very naturally. Hope this could help:)

``````/**
* 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; }
* }
*/
public class Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
return merge(intervals);
}
public List<Interval> merge(List<Interval> intervals) {
//1. sort the intervals by their starting points
//2. take the first interval and compare its end with the next interval starts:
//2a. if they overlap, update the end to be the max end of the overlapping intervals
//2b. if they don't overlap, add the previous "extended interval
if (intervals.size() <= 1) return intervals;
Collections.sort(intervals, new Comparator<Interval>() {
public int compare(Interval i1, Interval i2) {
return (i1.start - i2.start);
}
});

List<Interval> res = new ArrayList<>();
int start = intervals.get(0).start;
int end = intervals.get(0).end;
for (Interval curr : intervals) {
if (curr.start > end) { //no overlap //add
start = curr.start;
end = curr.end;
} else {//overlap //merge
end = Math.max(end, curr.end);
}
}