Merging Intervals

Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

Solution:

We first sort the list according to the start value. Then for any interval i in the middle, it has overlap with the next interval j if j.start <= i.end. If so, we know we are about to merge them into a new interval k. And k.start = i.start && k.end = max(i.end, j.end).

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;


public class MergeIntervals {
	
	public class Interval
	{
		public int start;
		public int end;
		
		public Interval(int start, int end) {
			super();
			this.start = start;
			this.end = end;
		}
	}
	
	public class MyComparator implements Comparator {
		public int compare(Object o1, Object o2) {
			Interval i1 = (Interval) o1;
			Interval i2 = (Interval) o2;
			return i1.start - i2.start;
		}
	}

	public ArrayList<Interval> merge(ArrayList<Interval> intervals) {
		if (intervals == null)
			return null;
		ArrayList<Interval> result = new ArrayList<Interval>();
		if (intervals.size() == 0)
			return result;
		if (intervals.size() == 1) {
			result.addAll(intervals);
			return result;
		}
		Collections.sort(intervals, new MyComparator());
		int start = intervals.get(0).start;
		int end = intervals.get(0).end;
		for (int i = 1; i < intervals.size(); ++i) {
			Interval curr = intervals.get(i);
			if (curr.start <= end) {
				end = Math.max(end, curr.end);
			} else {
				result.add(new Interval(start, end));
				start = curr.start;
				end = curr.end;
			}
		}
		result.add(new Interval(start, end));
		return result;
	}
}