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..
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!
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;
}
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。
@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)
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?