zoukankan      html  css  js  c++  java
  • 295. Find Median from Data Stream

    Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

    For example,

    [2,3,4], the median is 3

    [2,3], the median is (2 + 3) / 2 = 2.5

    Design a data structure that supports the following two operations:

    • void addNum(int num) - Add a integer number from the data stream to the data structure.
    • double findMedian() - Return the median of all elements so far.

    Example:

    addNum(1)
    addNum(2)
    findMedian() -> 1.5
    addNum(3) 
    findMedian() -> 2

    Follow up:

    1. If all integer numbers from the stream are between 0 and 100, how would you optimize it?
    2. If 99% of all integer numbers from the stream are between 0 and 100, how would you optimize it?

    用两个heap(min heap和max heap),较小的一半元素放进max heap,较大的一半元素放进min heap,两个heap的size之差不超1,最后的答案在两个堆顶元素中产生(min heap的最小值 -> 较大数里的最小值,max heap的最大值 -> 较小数里的最大值)。在平衡两个heap的大小时,如果大小之差=2了,把元素多的堆的堆顶元素弹出并加入到元素少的堆里。

    时间:O(logN) add element into heap + O(1) compute median -> O(NlogN),空间:O(N)

    class MedianFinder {
    
        /** initialize your data structure here. */
        PriorityQueue<Integer> minHeap;
        PriorityQueue<Integer> maxHeap;
        
        public MedianFinder() {
            minHeap = new PriorityQueue<>();
            maxHeap = new PriorityQueue<>((a, b) -> b - a);
        }
        
        public void addNum(int num) {
            if(maxHeap.isEmpty() || num <= maxHeap.peek())
                maxHeap.add(num);
            else
                minHeap.add(num);
            
            if(maxHeap.size() - minHeap.size() == 2)
                minHeap.add(maxHeap.poll());
            else if(minHeap.size() - maxHeap.size() == 2)
                maxHeap.add(minHeap.poll());
        }
        
        public double findMedian() {
            if(minHeap.size() > maxHeap.size())
                return (double)minHeap.peek();
            else if(maxHeap.size() > minHeap.size())
                return (double)maxHeap.peek();
            else
                return (double)(minHeap.peek() + maxHeap.peek()) / 2;
        }
    }
    
    /**
     * Your MedianFinder object will be instantiated and called as such:
     * MedianFinder obj = new MedianFinder();
     * obj.addNum(num);
     * double param_2 = obj.findMedian();
     */
  • 相关阅读:
    从太空到地球某个位置的轨迹录像制作 | Earth Zoom in/out Tutorial (Record Video)
    DNA replication timing数据库
    探索ENCODE数据库 | Encyclopedia of DNA Elements
    第100天: 三木板模型算法项目实战
    第99天:UDP 编程
    第98天:图像库 PIL 实例—验证码去噪
    第97天:图像库 PIL(二)
    第96天:图像库 PIL(一)
    第95天:StringIO & BytesIO
    第94天:数据分析之 pandas 初步
  • 原文地址:https://www.cnblogs.com/fatttcat/p/9998898.html
Copyright © 2011-2022 走看看