一个日期时间段有交集求并集的问题

同学工作中遇到的，问题大概是：输入一系列时间段（开始日期start，结束日期end）返回结果为：时间段若有交集则取其并集，无交集则返回原时间段，如输入：20160101，20160103；20160102，20160105；20160109，20160110；20160110，20160110；
则返回：20160101，20160105；20160109，20160110；
在这里我将输入的格式定为20160101,20160120格式代表开始和结束日期，并将其封装到Single对象中，对single对象初排序sort；然后再取并集，代码如下：

package calc;

import java.io.IOException;
import java.util.Collections;
import java.util.LinkedList;
import java.util.List;

public class Merge20170108 {
	public static void main(String[] args) throws IOException {
		List<Single> list = new LinkedList<Single>();
		Single s4 = new Single(new String[] { "20160302", "20160401" });
		Single s = new Single(new String[] { "20160101", "20160201" });
		Single s1 = new Single(new String[] { "20160111", "20160121" });
		Single s2 = new Single(new String[] { "20160301", "20160401" });
		Single s3 = new Single(new String[] { "20160301", "20160321" });
		list.add(s4);
		list.add(s);
		list.add(s1);
		list.add(s2);
		list.add(s3);
		merge(list);

	}

	// 对已经用开始日期start初排序的list进行交集合并处理
	public static void merge(List<Single> list) {
		Collections.sort(list, new Comparator());

		for (int i = 0; i < list.size() - 1; i++) {
			Single outer;
			Single inner;
			for (int j = i + 1; j < list.size(); j++) {
				outer = list.get(i);
				inner = list.get(j);
				String start = outer.start;
				String end = outer.end;
				String start2 = inner.start;
				String end2 = inner.end;
				if (end.compareTo(start2) >= 0 && end.compareTo(end2) <= 0) {
					outer.end = end2;
					list.set(i, outer);
					list.remove(j);
					j--;
				} else if (end.compareTo(end2) >= 0 || start.compareTo(end2) == 0) {
					list.remove(j);
					j--;
				}
			}
		}
		for (Single single : list) // 輸出操作
			System.out.println(single);
	}
}

class Single {
	String start;
	String end;

	public Single(String[] str) {
		start = str[0];
		end = str[1];
	}

	public String toString() {
		return start + "," + end;
	}

}

class Comparator implements java.util.Comparator<Single> {

	@Override
	public int compare(Single o1, Single o2) {
		return o1.start.compareTo(o2.start) > 0 ? 1 : -1;
	}

}