LeetCode-295. Find Median from Data Stream-优快云博客

本文介绍了一种高效的数据结构，用于处理数据流中的中位数计算问题。该数据结构支持添加整数并能返回所有元素的中位数。通过使用两个优先级队列（最小堆和最大堆），可以在O(log n)的时间复杂度内完成添加操作，并在O(1)时间内找到中位数。

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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:

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

class MedianFinder {

    int count ;
    PriorityQueue<Integer> minHeap;
    PriorityQueue<Integer> maxHeap;
    /** initialize your data structure here. */
    public MedianFinder() {
        count=0;
        minHeap = new PriorityQueue<>();
        maxHeap = new PriorityQueue<>(new Comparator<Integer>(){
            public int compare(Integer o1,Integer o2){
                return o2-o1;
            }
        });
    }
    
    public void addNum(int num) {
        if((count&1)==0){
            maxHeap.offer(num);
            int n = maxHeap.poll();
            minHeap.offer(n);
        }
        else{
            minHeap.offer(num);
            int n= minHeap.poll();
            maxHeap.offer(n);
        }
        count++;
    }
    
    public double findMedian() {
        if(count==0){
            return 0.0;
        }
        if((count&1)==0){
            return (minHeap.peek()+maxHeap.peek())/2.0;
        }
        else{
            return minHeap.peek()*1.0;
        }
    }
}