我正在解决一个难题,它涉及到分析所有大小为k的子集,并找出哪个是最优的。我写了一个解决方案,当子集的数量很少时,它可以工作,但对于更大的问题,它会耗尽内存。现在我正在尝试将一个用python编写的迭代函数翻译成java,这样我就可以在创建时分析每个子集,并且只获得代表它的优化程度的值,而不是整个子集,这样我就不会耗尽内存。这是我到目前为止所做的,即使是非常小的问题,它似乎也没有完成:
public static LinkedList<LinkedList<Integer>> getSets(int k, LinkedList<Integer> set)
{
int N = set.size();
int maxsets = nCr(N, k);
LinkedList<LinkedList<Integer>> toRet = new LinkedList<LinkedList<Integer>>();
int remains, thresh;
LinkedList<Integer> newset;
for (int i=0; i<maxsets; i++)
{
remains = k;
newset = new LinkedList<Integer>();
for (int val=1; val<=N; val++)
{
if (remains==0)
break;
thresh = nCr(N-val, remains-1);
if (i < thresh)
{
newset.add(set.get(val-1));
remains --;
}
else
{
i -= thresh;
}
}
toRet.add(newset);
}
return toRet;
}
有没有人可以帮我调试这个函数,或者提出另一个迭代生成k个子集的算法?
编辑:我终于让这个函数工作了,我必须创建一个与我相同的新变量来进行i和阈值比较,因为python处理循环索引的方式不同。
发布于 2015-05-18 15:00:58
您可以使用org.apache.commons.math3.util.Combinations。
示例:
import java.util.Arrays;
import java.util.Iterator;
import org.apache.commons.math3.util.Combinations;
public class tmp {
public static void main(String[] args) {
for (Iterator<int[]> iter = new Combinations(5, 3).iterator(); iter.hasNext();) {
System.out.println(Arrays.toString(iter.next()));
}
}
}
输出: 0,1,2 0,2,3 0,1,4 1,2,4 1,3,4
发布于 2013-02-13 15:50:45
这是我最近写的一个组合迭代器
package psychicpoker;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Iterator;
import java.util.List;
import static com.google.common.base.Preconditions.checkArgument;
public class CombinationIterator<T> implements Iterator<Collection<T>> {
private int[] indices;
private List<T> elements;
private boolean hasNext = true;
public CombinationIterator(List<T> elements, int k) throws IllegalArgumentException {
checkArgument(k<=elements.size(), "Impossible to select %d elements from hand of size %d", k, elements.size());
this.indices = new int[k];
for(int i=0; i<k; i++)
indices[i] = k-1-i;
this.elements = elements;
}
public boolean hasNext() {
return hasNext;
}
private int inc(int[] indices, int maxIndex, int depth) throws IllegalStateException {
if(depth == indices.length) {
throw new IllegalStateException("The End");
}
if(indices[depth] < maxIndex) {
indices[depth] = indices[depth]+1;
} else {
indices[depth] = inc(indices, maxIndex-1, depth+1)+1;
}
return indices[depth];
}
private boolean inc() {
try {
inc(indices, elements.size() - 1, 0);
return true;
} catch (IllegalStateException e) {
return false;
}
}
public Collection<T> next() {
Collection<T> result = new ArrayList<T>(indices.length);
for(int i=indices.length-1; i>=0; i--) {
result.add(elements.get(indices[i]));
}
hasNext = inc();
return result;
}
public void remove() {
throw new UnsupportedOperationException();
}
}
发布于 2013-06-03 18:01:34
灵感来自于afsantos的回答:-)...我决定编写一个C# .NET实现,从一个完整的集合中生成特定大小的所有子集组合。它不需要计算可能子集的总数;它会检测何时到达末尾。这就是它:
public static List<object[]> generateAllSubsetCombinations(object[] fullSet, ulong subsetSize) {
if (fullSet == null) {
throw new ArgumentException("Value cannot be null.", "fullSet");
}
else if (subsetSize < 1) {
throw new ArgumentException("Subset size must be 1 or greater.", "subsetSize");
}
else if ((ulong)fullSet.LongLength < subsetSize) {
throw new ArgumentException("Subset size cannot be greater than the total number of entries in the full set.", "subsetSize");
}
// All possible subsets will be stored here
List<object[]> allSubsets = new List<object[]>();
// Initialize current pick; will always be the leftmost consecutive x where x is subset size
ulong[] currentPick = new ulong[subsetSize];
for (ulong i = 0; i < subsetSize; i++) {
currentPick[i] = i;
}
while (true) {
// Add this subset's values to list of all subsets based on current pick
object[] subset = new object[subsetSize];
for (ulong i = 0; i < subsetSize; i++) {
subset[i] = fullSet[currentPick[i]];
}
allSubsets.Add(subset);
if (currentPick[0] + subsetSize >= (ulong)fullSet.LongLength) {
// Last pick must have been the final 3; end of subset generation
break;
}
// Update current pick for next subset
ulong shiftAfter = (ulong)currentPick.LongLength - 1;
bool loop;
do {
loop = false;
// Move current picker right
currentPick[shiftAfter]++;
// If we've gotten to the end of the full set, move left one picker
if (currentPick[shiftAfter] > (ulong)fullSet.LongLength - (subsetSize - shiftAfter)) {
if (shiftAfter > 0) {
shiftAfter--;
loop = true;
}
}
else {
// Update pickers to be consecutive
for (ulong i = shiftAfter+1; i < (ulong)currentPick.LongLength; i++) {
currentPick[i] = currentPick[i-1] + 1;
}
}
} while (loop);
}
return allSubsets;
}
https://stackoverflow.com/questions/4504974
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