Java for LeetCode 056 Merge 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].

解题思路一：

用两个指针startIndex和endIndex来维护每次添加intervals的start和end的位置，然后分类讨论即可，JAVA实现如下：

public List<Interval> merge(List<Interval> intervals) {
        List<Interval> list = new ArrayList<Interval>();
		if (intervals.size() == 0)
			return list;
		list.add(intervals.get(0));
		for (int i = 1; i < intervals.size(); i++) {
			Interval temp = intervals.get(i);
			int startIndex = 0, endIndex = 0;
			for (int j = 0; j < list.size(); j++) {
				if (temp.start > list.get(j).end) {
					startIndex += 2;
					endIndex += 2;
					continue;
				}
				if (temp.end < list.get(j).start)
					break;
				if (temp.start >= list.get(j).start)
					startIndex++;
				if (temp.end > list.get(j).end) {
					endIndex += 2;
					continue;
				}
				if (temp.end >= list.get(j).start)
					endIndex++;
				break;
			}
			if(startIndex==endIndex&&startIndex%2==0)
				list.add(startIndex/2,new Interval(temp.start,temp.end));
			else if(startIndex%2==0&&endIndex%2==0){
				list.get(startIndex/2).start=temp.start;
				list.get(startIndex/2).end=temp.end;
				for(int k=1;k<endIndex/2-startIndex/2;k++)
				list.remove(startIndex/2+1);
			}
			else if(startIndex%2==0&&endIndex%2!=0){
				list.get(startIndex/2).start=temp.start;
				list.get(startIndex/2).end=list.get(endIndex/2).end;
				for(int k=1;k<=endIndex/2-startIndex/2;k++)
					list.remove(startIndex/2+1);
			}
			else if(startIndex%2!=0&&endIndex%2==0){
				list.get(startIndex/2).end=temp.end;
				for(int k=1;k<endIndex/2-startIndex/2;k++)
					list.remove(startIndex/2+1);
			}
			else if(startIndex%2!=0&&endIndex%2!=0){
				list.get(startIndex/2).end=list.get(endIndex/2).end;
				for(int k=1;k<=endIndex/2-startIndex/2;k++)
					list.remove(startIndex/2+1);
			}
		}
		return list;
	}

 解题思路二：

首先构造一个比较器，对interval按照start进行排序，然后进行遍历，在遍历过程中，如果结果集合为空或者当前interval与结果集合中的最后一个interval不重叠，那么就直接将当前interval直接加入到结果集合中；如果发生了重叠，那么将结果集合的最后一个interval的右端点改为当前interval的右端点，JAVA实现如下：

	public List<Interval> merge(List<Interval> intervals) {
        List<Interval> list = new ArrayList<Interval>();
        Comparator<Interval> comparator = new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {
                if (o1.start == o2.start)
                    return o1.end - o2.end;
                return o1.start - o2.start;
            }
        };
        Collections.sort(intervals, comparator);
        for (Interval interval : intervals)
            if (list.size() == 0 || list.get(list.size() - 1).end < interval.start) 
                list.add(new Interval(interval.start, interval.end)); 
            else 
            	list.get(list.size() - 1).end = Math.max(interval.end, list.get(list.size() - 1).end);
        return list;
    }