42

像 Max-heap 和 Min-heap 一样,我想实现一个 Median-heap 来跟踪给定整数集的中位数。API 应具备以下三个功能:

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。这是到目前为止我所拥有的:中值堆将有两个整数来跟踪到目前为止插入的整数数量,它们是>当前中值(gcm)和<当前中值(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.

对于另一种情况也是如此。我想不出一个如何让元素下沉和游泳的算法。我认为应该考虑到数字与中位数的接近程度,例如:

private void swim(int k) {
    while (k > 1 && absless(k, k/2)) {   
        exch(k, k/2);
        k = k/2;
    }
}

不过,我无法提出整个解决方案。

4

6 回答 6

175

你需要两个堆:一个最小堆和一个最大堆。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。当最大堆比最小堆多一个元素时,中值在最大堆的顶部。

当两个堆包含相同数量的元素时,元素的总数是偶数。在这种情况下,您必须根据您对中位数的定义进行选择: a) 两个中间元素的平均值;b) 两者中的较大者;c) 较小的;d)随机选择两者中的任何一个......

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中值,则进入最小堆。如果它小于当前中值,则进入最大堆。然后你可能需要重新平衡。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/最大值并将其插入另一个堆。

为了构造元素列表的中值堆,我们应该首先使用线性时间算法并找到中值。一旦知道了中值,我们就可以根据中值简单地将元素添加到 Min-heap 和 Max-heap 中。不需要平衡堆,因为中位数会将输入的元素列表分成相等的两半。

如果您提取一个元素,您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变化。这样,您可以确保在任何时候,两个堆都具有相同的大小或仅相差一个元素。

于 2013-03-10T06:30:05.290 回答
11

这是一个 MedianHeap 的 java 实现,是在上面 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);
        }
    }
}
于 2016-01-14T16:28:58.640 回答
3

这里我的代码基于 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);
    }
  }
}
于 2017-03-04T04:36:50.697 回答
2

完美平衡的二叉搜索树 (BST) 不是中值堆吗?的确,即使是红黑 BST 也并不总是完美平衡,但它可能已经足够接近您的目的了。并且 log(n) 性能得到保证!

AVL 树比红黑 BST 更加平衡,因此它们更接近于真正的中值堆。

于 2013-03-10T09:47:26.887 回答
2

这是一个 Scala 实现,遵循上面 comocomocomocomo 的想法。

class MedianHeap(val capacity:Int) {
    private val minHeap = new PriorityQueue[Int](capacity / 2)
    private val maxHeap = new PriorityQueue[Int](capacity / 2, new Comparator[Int] {
      override def compare(o1: Int, o2: Int): Int = Integer.compare(o2, o1)
    })

    def add(x: Int): Unit = {
      if (x > median) {
        minHeap.add(x)
      } else {
        maxHeap.add(x)
      }

      // Re-balance the heaps.
      if (minHeap.size - maxHeap.size > 1) {
        maxHeap.add(minHeap.poll())
      }
      if (maxHeap.size - minHeap.size > 1) {
        minHeap.add(maxHeap.poll)
      }
    }

    def median: Double = {
      if (minHeap.isEmpty && maxHeap.isEmpty)
        return Int.MinValue
      if (minHeap.size == maxHeap.size) {
        return (minHeap.peek+ maxHeap.peek) / 2.0
      }
      if (minHeap.size > maxHeap.size) {
        return minHeap.peek()
      }
      maxHeap.peek
    }
  }
于 2016-10-31T08:27:07.977 回答
2

另一种不使用最大堆和最小堆的方法是立即使用中值堆。

在最大堆中,父级大于子级。我们可以有一种新类型的堆,其中父项位于子项的“中间”——左子项小于父项,右子项大于父项。所有偶数条目都是左孩子,所有奇数条目都是右孩子。

可以在最大堆中执行的相同游泳和下沉操作也可以在此中值堆中执行 - 稍作修改。在最大堆中的典型游泳操作中,插入的条目向上游泳直到小于父条目,在中值堆中,它将向上游泳直到小于父条目(如果它是奇数条目) 或大于父项(如果它是偶数项)。

这是我对这个中值堆的实现。为了简单起见,我使用了一个整数数组。

package priorityQueues;
import edu.princeton.cs.algs4.StdOut;

public class MedianInsertDelete {

    private Integer[] a;
    private int N;

    public MedianInsertDelete(int capacity){

        // accounts for '0' not being used
        this.a = new Integer[capacity+1]; 
        this.N = 0;
    }

    public void insert(int k){

        a[++N] = k;
        swim(N);
    }

    public int delMedian(){

        int median = findMedian();
        exch(1, N--);
        sink(1);
        a[N+1] = null;
        return median;

    }

    public int findMedian(){

        return a[1];


    }

    // entry swims up so that its left child is smaller and right is greater

    private void swim(int k){


        while(even(k) && k>1 && less(k/2,k)){

            exch(k, k/2);

            if ((N > k) && less (k+1, k/2)) exch(k+1, k/2);
            k = k/2;
        }

        while(!even(k) && (k>1 && !less(k/2,k))){

            exch(k, k/2);
            if (!less (k-1, k/2)) exch(k-1, k/2);
            k = k/2;
        }

    }

// if the left child is larger or if the right child is smaller, the entry sinks down
    private void sink (int k){

        while(2*k <= N){
            int j = 2*k;
            if (j < N && less (j, k)) j++;
            if (less(k,j)) break;
            exch(k, j);
            k = j;
        }

    }

    private boolean even(int i){

        if ((i%2) == 0) return true;
        else return false;
    }

    private void exch(int i, int j){

        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }

    private boolean less(int i, int j){

        if (a[i] <= a[j]) return true;
        else return false;
    }


    public static void main(String[] args) {

        MedianInsertDelete medianInsertDelete = new MedianInsertDelete(10);

        for(int i = 1; i <=10; i++){

            medianInsertDelete.insert(i);
        }

        StdOut.println("The median is: " + medianInsertDelete.findMedian());

        medianInsertDelete.delMedian();


        StdOut.println("Original median deleted. The new median is " + medianInsertDelete.findMedian());




    }
}


于 2018-11-29T01:37:26.063 回答