间隔问题，合并间隔(merge interval)，插入间隔(insert interval)

Merge Interval:

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].

算法分析：首先要对给定序列排序。然后再去遍历合并。

class Interval
{
	int start;
	int end;
	Interval()
	{
		start = 0;
		end = 0;
	}
	Interval(int s, int e)
	{
		start = s;
		end = e;
	}
}
public class MergeIntervals
{
	public List<Interval> merge(List<Interval> intervals)
	{
		List<Interval> res = new ArrayList<>();
		if(intervals == null || intervals.size() == 0)
		{
			return res;
		}
		Collections.sort(intervals, new Comparator<Interval>()//自定义比较方法，外部实现Comparator接口
				{
					public int compare(Interval i1, Interval i2)
					{
						if(i1.start != i2.start)
						{
							return i1.start - i2.start;
						}
						else
						{
							return i1.end - i2.end;
						}
					}
				});
		Interval pre = intervals.get(0);
		for(int i = 0; i < intervals.size(); i ++)
		{
			Interval curr = intervals.get(i);
			if(curr.start > pre.end)
			{
				res.add(pre);
				pre = curr;
			}
			else
			{
				Interval merged = new Interval(pre.start, Math.max(pre.end, curr.end));
				pre = merged;
			}
		}
		res.add(pre);
		return res;
	}
}

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].

给定间隔序列和插入间隔，默认给定间隔序列有序，返回合并后的不重叠的间隔序列。是上一道题的变形。

public class InsertInterval
{
	public List<Interval> insert(List<Interval> intervals, Interval newInterval)
	{
		List<Interval> res = new ArrayList<>();
		for (Interval interval : intervals)
		{
			if(interval.end < newInterval.start)
			{
				res.add(interval);
			}
			else if(interval.start > newInterval.end)
			{
				res.add(newInterval);
				newInterval = interval;
			}
			else if(interval.start <= newInterval.end || interval.end >= newInterval.start)
			{
				newInterval = new Interval(Math.min(interval.start, newInterval.start),Math.max(interval.end, newInterval.end));
			}
		}
		res.add(newInterval);
		return res;
    }
}