有效地求出随机序列的中值

Question

数字是随机生成的，并传递给一个方法。编写一个程序，在生成新值时查找和维护中位数。

堆的大小可以相等，或者下面的堆有一个额外的大小。

private Comparator<Integer> maxHeapComparator, minHeapComparator;
private PriorityQueue<Integer> maxHeap, minHeap;

public void addNewNumber(int randomNumber) {
  if (maxHeap.size() == minHeap.size()) {
    if ((minHeap.peek() != null) && randomNumber > minHeap.peek()) {
      maxHeap.offer(minHeap.poll());
      minHeap.offer(randomNumber);
    } else {
      maxHeap.offer(randomNumber);
    }
  }
  else {  // why the following block is correct? 
    // I think it may create unbalanced heap size
    if(randomNumber < maxHeap.peek()) {
      minHeap.offer(maxHeap.poll());
      maxHeap.offer(randomNumber);
    }
    else {
      minHeap.offer(randomNumber);
    }
  }
}

public static double getMedian() {
  if (maxHeap.isEmpty()) return minHeap.peek();
  else if (minHeap.isEmpty()) return maxHeap.peek();

  if (maxHeap.size() == minHeap.size()) {
    return (minHeap.peek() + maxHeap.peek()) / 2;
  } else if (maxHeap.size() > minHeap.size()) {
    return maxHeap.peek();
  } else {
    return minHeap.peek();
  }
}

假设解决方案是正确的，那么我就不明白为什么代码块(参见我的注释)可以保持堆大小平衡。换句话说，两个堆的大小之差是0或1。

Let us see an example, given a sequence 1, 2, 3, 4, 5
The first random number is **1**
    max-heap: 1
    min-heap:

The second random number is **2**
    max-heap: 1
    min-heap: 2

The third random number is **3**
    max-heap: 1 2
    min-heap: 3 4

The fourth random number is **4**
    max-heap: 1 2 3
    min-heap: 4 5

谢谢

Answer 1

在通过给定的序列运行它之后，

max-heap : 1, 2, 3
min-heap : 4, 5

因为max-heap大小大于min-heap，所以它返回3作为中间值。

max-heap大约存储左半个元素，min-heap大约存储右半个序列。

这段代码偏向于最大堆的左半部分。

我不明白为什么这个代码是不正确的。