Merge Intervals

//Merge Intervals Total Accepted: 40697 Total Submissions: 181352 My Submissions Question Solution 
//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].
这道题真的是眨眼间的灵感，一开始没想到先拍个序，一直去用一个数去和现在所有存在的遍历，就算遍历完了，还要有多个还是合并，但是直接拍完序就可以直接来对最后一个处理了，使用stack就可以了。
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Currency;
import java.util.List;
import java.util.Stack;


class Interval {
     int start;
     int end;
     Interval() { start = 0; end = 0; }
     Interval(int s, int e) { start = s; end = e; }
 }
public class MergeIntervals {


public static void main(String[] args) {
// TODO Auto-generated method stub


}


@SuppressWarnings("unchecked")
public List<Interval> merge(List<Interval> intervals) {
  if(intervals==null||intervals.size()<=1)
  return intervals;
  
  Collections.sort(intervals,new Comparator() {
     @Override
     public int compare(Object o1, Object o2) {
       return ((Interval)o1).start-((Interval)o2).start;
     }
   });
  Stack<Interval> stack=new Stack<>();
  stack.add(intervals.get(0));
  for(int i=1;i<intervals.size();i++){
  Interval cur=stack.pop();
  if(intervals.get(i).start>cur.end){
  stack.push(cur);
  stack.push(intervals.get(i));
  }else{
  cur.end=Math.max(cur.end, intervals.get(i).end);
  stack.push(cur);
  }
  }
  List<Interval> result=new ArrayList<>();
  while(!stack.isEmpty()){
  result.add(stack.pop());
  }
  return result;
}
}