1
数字是随机生成的并传递给一个方法。编写一个程序来查找和维护生成新值的中值。有效地找到随机序列的中值
堆大小可以相等,或者下面的堆有一个额外的堆。
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();
}
}
假设该解决方案是正确的,那么我不明白为什么代码块(见我的意见)可以保持堆大小平衡。换句话说,两个堆的尺寸差为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
Eww行号。还标记语法突出显示的正确语言。 – Joe 2011-04-14 20:34:17
这功课吗?如果是这样,相应标记。 – Leonel 2011-04-14 20:42:04
“写一个程序..”忘记,这是**你的**作业,而不是我们的。 – 2011-04-14 21:16:49