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].
本题使用小顶堆来做，将interval.start进行排序，然后取出第一个interval值，起始点设为start，终止点设为end，然后看下一个元素的起始点是否在这个start和end之间，如果在的话，就看它们的end哪个大，把大的存为end，如果起始点不在这个范围内，则把start，end存入res里面，代码如下：
/**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
public class Solution {
public List<Interval> merge(List<Interval> intervals) {
if(intervals.size()<=1) return intervals;
List<Interval> res = new ArrayList<>();
PriorityQueue<Interval> q = new PriorityQueue<Interval>(intervals.size(),new Comparator<Interval>(){
public int compare(Interval a,Interval b){
return a.start-b.start;
}
});
for(Interval i:intervals){
q.offer(i);
}
Interval val = q.poll();
int start = val.start;
int end = val.end;
while(!q.isEmpty()){
Interval i = q.poll();
if(i.start<=end){
end = Math.max(end,i.end);
}else{
res.add(new Interval(start,end));
start = i.start;
end = i.end;
}
}
res.add(new Interval(start,end));
return res;
}
}
原文：http://www.cnblogs.com/codeskiller/p/6362250.html