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算法1：
1、利用ArrayList维持一个有序队列
2、取ArrayList中位数

import java.util.*;
public class Solution {
    //第一版  每次插入都用插入排序来排序
    List<Integer> list = new ArrayList<>();
    public void Insert(Integer num) {
          if(list.isEmpty()) {
            list.add(num);
            return;
        }
        for(int i=list.size()-1;i>=0;i--){
            if(list.get(i)<=num){
                list.add(i+1,num);
                break;
            }else if(i==0){
                list.add(0,num);
                break;
            }
        }
    }

    public Double GetMedian() {
        if(list.size()==1) return list.get(0)+0d;

        int index = list.size()/2;
        if(list.size()%2==0){
            //偶数个
            return  (list.get(index-1)+list.get(index))/2d;
        }else{
            //奇数个
            return list.get(index)+0d;
        }


    }


}

算法2：利用大根堆小根堆配合实现（有点绕）

import java.util.*;
public class Solution {
    //用大根堆和小根堆来配合实现,想不清楚的用几个数据演练一下就明白了
    //小根堆 8765
    private PriorityQueue<Integer> minHeap = new PriorityQueue<>();
    //大根堆 1234       
    private PriorityQueue<Integer> maxHeap = new PriorityQueue<>(new Comparator<Integer>(){
        public int compare(Integer o1,Integer o2){
            return o2-o1;
        }
    });
    //判断奇数偶数
    int count = 0;
    public void Insert(Integer num) {
         if(count%2==0){
             maxHeap.offer(num);
             minHeap.offer(maxHeap.poll());
         }else{
             minHeap.offer(num);
             maxHeap.offer(minHeap.poll());
         }
        count++;
    }
    public Double GetMedian() {
        if(count%2==0){
            return (minHeap.peek()+maxHeap.peek())/2.0;
        }else{
            return minHeap.peek()+0d;
        }

    }


}