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
     
       merge(List
      
        intervals) {        List
       
         list = new ArrayList
        
         ();		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
        
       
      
     

 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
     
       merge(List
      
        intervals) {        List
       
         list = new ArrayList
        
         ();        Comparator
         
           comparator = new Comparator
          
           () { @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; }