我正在尝试用Java实现Median Median of Medians:
Select(Comparable[] list, int pos, int colSize, int colMed)
list
是要查找指定位置的值列表pos
是指定的位置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,或者如何准确地实现它。
答案 0 :(得分:9)
我不知道您是否仍然需要解决此问题,但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!
答案 1 :(得分:4)
问的问题是Java,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,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,上述代码实际使用速度太慢。
这是一种更简单的方法来获得不保证性能的第k个元素,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;
}
答案 2 :(得分:2)
我同意Chip Uni的答案/解决方案。我将只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,区别在于只有一个分区被解决(左或右)。我们只需找到一个最佳枢轴,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O(N的时间复杂度) )。上述算法称为中位数中位数,计算中位数为5的中位数,结果表明算法具有线性时间复杂度。
然而,当搜索范围为第n个最小/最大元素(我想你用这个算法实现)时,使用排序算法来加速算法。插入排序在最多7到10个元素的小阵列上特别快。
实施说明:
M = select({x[i]}, n/10)
实际上意味着取5个元素组的所有中位数的中位数。您可以通过创建另一个大小为(n - 1)/5 + 1
的数组来完成此操作,并递归调用相同的算法来查找第n / 10个元素(这是新创建的数组的中位数)。
答案 3 :(得分:0)
@android开发者:
for (i = 1 to n/5) do
x[i] = select(S[i],3)
真的是
for (i = 1 to ceiling(n/5) do
x[i] = select(S[i],3)
具有适合您数据的天花板功能(例如,在java 2双打中) 这会影响中位数以及简单地取n / 10,但我们发现最接近阵列中出现的均值,而不是真正的均值。 另一个注意事项是S [i]可能少于3个元素,因此我们希望找到相对于长度的中位数;将它传递给选择,k = 3,总是不行。(例如,n = 11,我们有3个子组2 w 5,1 w 1元素)
答案 4 :(得分:-1)
我知道这是一篇非常古老的帖子,你可能不记得了。但我想知道你实施它时是否衡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,当数组的大小大约是十万或更多元素时,这种算法只能胜过java排序算法。并且它只快2到3倍,显然不是log(n)时间更快。
你对此有何评论?