Insert Interval

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].

Solution

/**
 * 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> merge(List<Interval> intervals) {
        List<Interval> result = new ArrayList<>();
        if (intervals == null || intervals.size() == 0) {
            return result;
        }

        intervals.sort(new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {
                if (o1.start < o2.start) {
                    return -1;
                } else if (o1.start == o2.start) {
                    return 0;
                } else {
                    return 1;
                }
            }
        });

        int start = intervals.get(0).start;
        int end = intervals.get(0).end;

        for (int i = 1; i < intervals.size(); i++) {
            Interval currentInterval = intervals.get(i);
            if (end < currentInterval.start) {
                result.add(new Interval(start, end));
                start = currentInterval.start;
                end = currentInterval.end;
            } else {
                end = currentInterval.end < end ? end: currentInterval.end;
            }
        }

        result.add(new Interval(start, end));
        return result;
    }

    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        intervals.add(newInterval);
        return merge(intervals);
    }
}