2

我正在尝试在 Java 中为这样的方法实现 Median of Median:

Select(Comparable[] list, int pos, int colSize, int colMed)
  • list是要查找指定位置的值的列表
  • pos是指定位置
  • colSize是我在第一阶段创建的列的大小
  • colMed是我用作 medX 的那些列中的位置

我不确定哪种排序算法最适合使用或如何准确实现它..

4

5 回答 5

11

我不知道你是否还需要解决这个问题,但是http://www.ics.uci.edu/~eppstein/161/960130.html有一个算法:

select(L,k)
{
    if (L has 10 or fewer elements)
    {
        sort L
        return the element in the kth position
    }

    partition L into subsets S[i] of five elements each
        (there will be n/5 subsets total).

    for (i = 1 to n/5) do
        x[i] = select(S[i],3)

    M = select({x[i]}, n/10)

    partition L into L1<M, L2=M, L3>M
    if (k <= length(L1))
        return select(L1,k)
    else if (k > length(L1)+length(L2))
        return select(L3,k-length(L1)-length(L2))
    else return M
}

祝你好运!

于 2009-12-07T05:50:37.557 回答
5

这个问题是针对 Java 的,所以这里是

import java.util.*;

public class MedianOfMedians {
    private MedianOfMedians() {

    }

    /**
     * Returns median of list in linear time.
     * 
     * @param list list to search, which may be reordered on return
     * @return median of array in linear time.
     */
    public static Comparable getMedian(ArrayList<Comparable> list) {
        int s = list.size();
        if (s < 1)
            throw new IllegalArgumentException();
        int pos = select(list, 0, s, s / 2);
        return list.get(pos);
    }

    /**
     * Returns position of k'th largest element of sub-list.
     * 
     * @param list list to search, whose sub-list may be shuffled before
     *            returning
     * @param lo first element of sub-list in list
     * @param hi just after last element of sub-list in list
     * @param k
     * @return position of k'th largest element of (possibly shuffled) sub-list.
     */
    public static int select(ArrayList<Comparable> list, int lo, int hi, int k) {
        if (lo >= hi || k < 0 || lo + k >= hi)
            throw new IllegalArgumentException();
        if (hi - lo < 10) {
            Collections.sort(list.subList(lo, hi));
            return lo + k;
        }
        int s = hi - lo;
        int np = s / 5; // Number of partitions
        for (int i = 0; i < np; i++) {
            // For each partition, move its median to front of our sublist
            int lo2 = lo + i * 5;
            int hi2 = (i + 1 == np) ? hi : (lo2 + 5);
            int pos = select(list, lo2, hi2, 2);
            Collections.swap(list, pos, lo + i);
        }

        // Partition medians were moved to front, so we can recurse without making another list.
        int pos = select(list, lo, lo + np, np / 2);

        // Re-partition list to [<pivot][pivot][>pivot]
        int m = triage(list, lo, hi, pos);
        int cmp = lo + k - m;
        if (cmp > 0)
            return select(list, m + 1, hi, k - (m - lo) - 1);
        else if (cmp < 0)
            return select(list, lo, m, k);
        return lo + k;
    }

    /**
     * Partition sub-list into 3 parts [<pivot][pivot][>pivot].
     * 
     * @param list
     * @param lo
     * @param hi
     * @param pos input position of pivot value
     * @return output position of pivot value
     */
    private static int triage(ArrayList<Comparable> list, int lo, int hi,
            int pos) {
        Comparable pivot = list.get(pos);
        int lo3 = lo;
        int hi3 = hi;
        while (lo3 < hi3) {
            Comparable e = list.get(lo3);
            int cmp = e.compareTo(pivot);
            if (cmp < 0)
                lo3++;
            else if (cmp > 0)
                Collections.swap(list, lo3, --hi3);
            else {
                while (hi3 > lo3 + 1) {
                    assert (list.get(lo3).compareTo(pivot) == 0);
                    e = list.get(--hi3);
                    cmp = e.compareTo(pivot);
                    if (cmp <= 0) {
                        if (lo3 + 1 == hi3) {
                            Collections.swap(list, lo3, lo3 + 1);
                            lo3++;
                            break;
                        }
                        Collections.swap(list, lo3, lo3 + 1);
                        assert (list.get(lo3 + 1).compareTo(pivot) == 0);
                        Collections.swap(list, lo3, hi3);
                        lo3++;
                        hi3++;
                    }
                }
                break;
            }
        }
        assert (list.get(lo3).compareTo(pivot) == 0);
        return lo3;
    }

}

这是一个单元测试来检查它是否有效......

import java.util.*;

import junit.framework.TestCase;

public class MedianOfMedianTest extends TestCase {
    public void testMedianOfMedianTest() {
        Random r = new Random(1);
        int n = 87;
        for (int trial = 0; trial < 1000; trial++) {
            ArrayList list = new ArrayList();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                int v = r.nextInt(256);
                a[i] = v;
                list.add(v);
            }
            int m1 = (Integer)MedianOfMedians.getMedian(list);
            Arrays.sort(a);
            int m2 = a[n/2];
            assertEquals(m1, m2);
        }
    }
}

但是,上面的代码对于实际使用来说太慢了。

这是获取第 k 个元素的更简单方法,它不能保证性能,但在实践中要快得多:

/**
 * Returns position of k'th largest element of sub-list.
 * 
 * @param list list to search, whose sub-list may be shuffled before
 *            returning
 * @param lo first element of sub-list in list
 * @param hi just after last element of sub-list in list
 * @param k
 * @return position of k'th largest element of (possibly shuffled) sub-list.
 */
static int select(double[] list, int lo, int hi, int k) {
    int n = hi - lo;
    if (n < 2)
        return lo;

    double pivot = list[lo + (k * 7919) % n]; // Pick a random pivot

    // Triage list to [<pivot][=pivot][>pivot]
    int nLess = 0, nSame = 0, nMore = 0;
    int lo3 = lo;
    int hi3 = hi;
    while (lo3 < hi3) {
        double e = list[lo3];
        int cmp = compare(e, pivot);
        if (cmp < 0) {
            nLess++;
            lo3++;
        } else if (cmp > 0) {
            swap(list, lo3, --hi3);
            if (nSame > 0)
                swap(list, hi3, hi3 + nSame);
            nMore++;
        } else {
            nSame++;
            swap(list, lo3, --hi3);
        }
    }
    assert (nSame > 0);
    assert (nLess + nSame + nMore == n);
    assert (list[lo + nLess] == pivot);
    assert (list[hi - nMore - 1] == pivot);
    if (k >= n - nMore)
        return select(list, hi - nMore, hi, k - nLess - nSame);
    else if (k < nLess)
        return select(list, lo, lo + nLess, k);
    return lo + k;
}
于 2014-12-31T10:27:16.380 回答
2

我同意 Chip Uni 的回答/解决方案。我将仅评论排序部分并提供一些进一步的解释:

您不需要任何排序算法。该算法类似于快速排序,不同之处在于只解决了一个分区(左或右)。我们只需要找到一个最佳枢轴,使左右部分尽可能相等,这意味着 N/2 + N/4 + N/8 ... = 2N 次迭代,因此时间复杂度为 O(N )。上述算法称为中位数的中位数,计算中位数为 5 的中位数,结果得出算法的线性时间复杂度。

但是,在搜索范围内的第 n 个最小/最大元素(我想您正在使用此算法实现)时使用排序算法以加快算法速度。插入排序在多达 7 到 10 个元素的小型数组上特别快。

实施说明:

M = select({x[i]}, n/10)

实际上意味着取 5 元素组的所有这些中位数的中位数。您可以通过创建另一个大小的数组(n - 1)/5 + 1并递归调用相同的算法来找到第 n/10 个元素(这是新创建的数组的中位数)来实现这一点。

于 2011-02-04T11:49:02.847 回答
0

@android 开发者:

for (i = 1 to n/5) do
    x[i] = select(S[i],3)

是真的

for (i = 1 to ceiling(n/5) do
    x[i] = select(S[i],3)

具有适合您的数据的上限函数(例如在 java 2 双打中) 这会影响中位数以及简单地取 n/10,但我们发现最接近数组中出现的平均值,而不是真实平均值。另一个注意事项是 S[i] 的元素可能少于 3 个,因此我们要找到长度的中位数;将它传递给 k=3 的 select 并不总是有效。(例如 n =11,我们有 3 个子组 2 w 5, 1 w 1 元素)

于 2016-02-06T23:33:12.863 回答
-1

我知道这是一个非常古老的帖子,您可能不再记得它了。但是我想知道您在实施时是否测量了实施的运行时间?

我尝试了这个算法并将其与使用 java 排序方法 (Arrays.sort() ) 的简单方法进行比较,然后从排序数组中选择第 k 个元素。我收到的结果是,该算法仅在数组大小约为十万个元素或更多时才胜过 java 排序算法。而且它只快了大约 2 到 3 倍,这显然不是 log(n) 时间快。

你对此有何评论?

于 2011-09-27T05:27:55.953 回答