Cracking the coding interview--Q20.9

题目

原文：

Numbers are randomly generated and passed to a method. Write a program to find and maintain the median value as new values are generated.

译文：

随机产生一些数据，并传递给一个方法，写一个程序找出并维护这些数的中位数。

解答

ctci解法：

一个解决的方法是使用两个优先堆：一个最大堆存储比中位数小一点的值，最小堆存储比中位数大的值，这样中位数就是最大堆的最大值；当一个新的的值到达时，如果它小于或等于中位值则放在堆的下面，否则放在堆的上面。这堆的大小是等于小于堆的额外大小。这样容易通过从一个堆到另一个堆上移元素来被存储。这样中位值是可以达到的更新时间为O(lgn)

代码如下：

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 {
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();
}
}

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