像Max-heap和Min-heap一样,我想实现一个中位数堆来跟踪一组给定整数的中位数。API需要有以下三个函数:
insert(int) // should take O(logN)
int median() // will be the topmost element of the heap. O(1)
int delmedian() // should take O(logN)我想使用数组(a)实现来实现堆,其中数组索引k的子项存储在数组索引2*k和2*k + 1中。为方便起见,数组从索引1开始填充元素。这是我到目前为止所做的:中位数堆将有两个整数来跟踪到目前为止插入的整数的数量,即>当前中位数(gcm)和<当前中位数(lcm)。
if abs(gcm-lcm) >= 2 and gcm > lcm we need to swap a[1] with one of its children.
The child chosen should be greater than a[1]. If both are greater,
choose the smaller of two.另一种情况也是如此。我想不出如何沉没和游动元素的算法。我认为它应该考虑到数字与中位数的接近程度,比如:
private void swim(int k) {
while (k > 1 && absless(k, k/2)) {
exch(k, k/2);
k = k/2;
}
}不过,我想不出完整的解决方案。
发布于 2013-03-10 14:30:05
您需要两个堆:一个min-heap和一个max-heap。每个堆包含大约一半的数据。最小堆中的每个元素都大于或等于中位数,最大堆中的每个元素都小于或等于中位数。
当min-heap比max-heap多包含一个元素时,中位数位于min-heap的顶部。当max-heap比min-heap多包含一个元素时,中位数位于max-heap的顶部。
当两个堆包含相同数量的元素时,元素总数为偶数。在这种情况下,你必须根据中位数的定义进行选择: a)两个中间元素的平均值;b)两个元素中较大的一个;c)较小的;d)随机选择两个元素中的任何一个...
每次插入时,将新元素与堆顶部的元素进行比较,以确定将其插入的位置。如果新元素大于当前的中位数,它将转到min-heap。如果它小于当前的中位数,它将转到最大堆。然后,您可能需要重新平衡。如果堆的大小相差不止一个元素,则从具有更多元素的堆中提取最小/最大值,并将其插入到另一个堆中。
为了构造元素列表的中位数堆,我们应该首先使用线性时间算法并找到中位数。一旦知道了中位数,我们就可以简单地根据中位数向Min-heap和Max-heap添加元素。平衡堆并不是必需的,因为中位数会将输入元素列表分成相等的两部分。
如果提取一个元素,可能需要通过将一个元素从一个堆移动到另一个堆来补偿大小的变化。这样可以确保在任何时候,两个堆都具有相同的大小或只有一个元素不同。
发布于 2016-01-15 00:28:59
这里是一个MedianHeap的java实现,它是在上述comocomocomocomo的解释的帮助下开发的。
import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;
/**
*
* @author BatmanLost
*/
public class MedianHeap {
//stores all the numbers less than the current median in a maxheap, i.e median is the maximum, at the root
private PriorityQueue<Integer> maxheap;
//stores all the numbers greater than the current median in a minheap, i.e median is the minimum, at the root
private PriorityQueue<Integer> minheap;
//comparators for PriorityQueue
private static final maxHeapComparator myMaxHeapComparator = new maxHeapComparator();
private static final minHeapComparator myMinHeapComparator = new minHeapComparator();
/**
* Comparator for the minHeap, smallest number has the highest priority, natural ordering
*/
private static class minHeapComparator implements Comparator<Integer>{
@Override
public int compare(Integer i, Integer j) {
return i>j ? 1 : i==j ? 0 : -1 ;
}
}
/**
* Comparator for the maxHeap, largest number has the highest priority
*/
private static class maxHeapComparator implements Comparator<Integer>{
// opposite to minHeapComparator, invert the return values
@Override
public int compare(Integer i, Integer j) {
return i>j ? -1 : i==j ? 0 : 1 ;
}
}
/**
* Constructor for a MedianHeap, to dynamically generate median.
*/
public MedianHeap(){
// initialize maxheap and minheap with appropriate comparators
maxheap = new PriorityQueue<Integer>(11,myMaxHeapComparator);
minheap = new PriorityQueue<Integer>(11,myMinHeapComparator);
}
/**
* Returns empty if no median i.e, no input
* @return
*/
private boolean isEmpty(){
return maxheap.size() == 0 && minheap.size() == 0 ;
}
/**
* Inserts into MedianHeap to update the median accordingly
* @param n
*/
public void insert(int n){
// initialize if empty
if(isEmpty()){ minheap.add(n);}
else{
//add to the appropriate heap
// if n is less than or equal to current median, add to maxheap
if(Double.compare(n, median()) <= 0){maxheap.add(n);}
// if n is greater than current median, add to min heap
else{minheap.add(n);}
}
// fix the chaos, if any imbalance occurs in the heap sizes
//i.e, absolute difference of sizes is greater than one.
fixChaos();
}
/**
* Re-balances the heap sizes
*/
private void fixChaos(){
//if sizes of heaps differ by 2, then it's a chaos, since median must be the middle element
if( Math.abs( maxheap.size() - minheap.size()) > 1){
//check which one is the culprit and take action by kicking out the root from culprit into victim
if(maxheap.size() > minheap.size()){
minheap.add(maxheap.poll());
}
else{ maxheap.add(minheap.poll());}
}
}
/**
* returns the median of the numbers encountered so far
* @return
*/
public double median(){
//if total size(no. of elements entered) is even, then median iss the average of the 2 middle elements
//i.e, average of the root's of the heaps.
if( maxheap.size() == minheap.size()) {
return ((double)maxheap.peek() + (double)minheap.peek())/2 ;
}
//else median is middle element, i.e, root of the heap with one element more
else if (maxheap.size() > minheap.size()){ return (double)maxheap.peek();}
else{ return (double)minheap.peek();}
}
/**
* String representation of the numbers and median
* @return
*/
public String toString(){
StringBuilder sb = new StringBuilder();
sb.append("\n Median for the numbers : " );
for(int i: maxheap){sb.append(" "+i); }
for(int i: minheap){sb.append(" "+i); }
sb.append(" is " + median()+"\n");
return sb.toString();
}
/**
* Adds all the array elements and returns the median.
* @param array
* @return
*/
public double addArray(int[] array){
for(int i=0; i<array.length ;i++){
insert(array[i]);
}
return median();
}
/**
* Just a test
* @param N
*/
public void test(int N){
int[] array = InputGenerator.randomArray(N);
System.out.println("Input array: \n"+Arrays.toString(array));
addArray(array);
System.out.println("Computed Median is :" + median());
Arrays.sort(array);
System.out.println("Sorted array: \n"+Arrays.toString(array));
if(N%2==0){ System.out.println("Calculated Median is :" + (array[N/2] + array[(N/2)-1])/2.0);}
else{System.out.println("Calculated Median is :" + array[N/2] +"\n");}
}
/**
* Another testing utility
*/
public void printInternal(){
System.out.println("Less than median, max heap:" + maxheap);
System.out.println("Greater than median, min heap:" + minheap);
}
//Inner class to generate input for basic testing
private static class InputGenerator {
public static int[] orderedArray(int N){
int[] array = new int[N];
for(int i=0; i<N; i++){
array[i] = i;
}
return array;
}
public static int[] randomArray(int N){
int[] array = new int[N];
for(int i=0; i<N; i++){
array[i] = (int)(Math.random()*N*N);
}
return array;
}
public static int readInt(String s){
System.out.println(s);
Scanner sc = new Scanner(System.in);
return sc.nextInt();
}
}
public static void main(String[] args){
System.out.println("You got to stop the program MANUALLY!!");
while(true){
MedianHeap testObj = new MedianHeap();
testObj.test(InputGenerator.readInt("Enter size of the array:"));
System.out.println(testObj);
}
}
}发布于 2017-03-04 12:36:51
下面是我的代码,基于comocomocomocomo提供的答案:
import java.util.PriorityQueue;
public class Median {
private PriorityQueue<Integer> minHeap =
new PriorityQueue<Integer>();
private PriorityQueue<Integer> maxHeap =
new PriorityQueue<Integer>((o1,o2)-> o2-o1);
public float median() {
int minSize = minHeap.size();
int maxSize = maxHeap.size();
if (minSize == 0 && maxSize == 0) {
return 0;
}
if (minSize > maxSize) {
return minHeap.peek();
}if (minSize < maxSize) {
return maxHeap.peek();
}
return (minHeap.peek()+maxHeap.peek())/2F;
}
public void insert(int element) {
float median = median();
if (element > median) {
minHeap.offer(element);
} else {
maxHeap.offer(element);
}
balanceHeap();
}
private void balanceHeap() {
int minSize = minHeap.size();
int maxSize = maxHeap.size();
int tmp = 0;
if (minSize > maxSize + 1) {
tmp = minHeap.poll();
maxHeap.offer(tmp);
}
if (maxSize > minSize + 1) {
tmp = maxHeap.poll();
minHeap.offer(tmp);
}
}
}https://stackoverflow.com/questions/15319561
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