我有以下地图:Map<Integer,String[]> map = new HashMap<Integer,String[]>();
键是整数,值是数组(也可以用列表替换)。
现在,我希望获得键之间所有可能的值组合。例如,假设地图包含以下条目:
key 1: "test1", "stackoverflow"
key 2: "test2", "wow"
key 3: "new"
组合由
组成("test1","test2","new")
("test1","wow","new")
("stackoverflow", "test2", "new")
("stackoverflow", "wow", "new")
为此我想象一个方法boolean hasNext()
,如果有下一对,则返回true,第二个方法只返回下一组值(如果有的话)。
如何做到这一点?地图也可以用其他数据结构代替。
答案 0 :(得分:8)
算法基本上与十进制数的增量算法(“x - &gt; x + 1”)几乎相同。
这里是迭代器类:
import java.util.Iterator;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.TreeSet;
public class CombinationsIterator implements Iterator<String[]> {
// Immutable fields
private final int combinationLength;
private final String[][] values;
private final int[] maxIndexes;
// Mutable fields
private final int[] currentIndexes;
private boolean hasNext;
public CombinationsIterator(final Map<Integer,String[]> map) {
combinationLength = map.size();
values = new String[combinationLength][];
maxIndexes = new int[combinationLength];
currentIndexes = new int[combinationLength];
if (combinationLength == 0) {
hasNext = false;
return;
}
hasNext = true;
// Reorganize the map to array.
// Map is not actually needed and would unnecessarily complicate the algorithm.
int valuesIndex = 0;
for (final int key : new TreeSet<>(map.keySet())) {
values[valuesIndex++] = map.get(key);
}
// Fill in the arrays of max indexes and current indexes.
for (int i = 0; i < combinationLength; ++i) {
if (values[i].length == 0) {
// Set hasNext to false if at least one of the value-arrays is empty.
// Stop the loop as the behavior of the iterator is already defined in this case:
// the iterator will just return no combinations.
hasNext = false;
return;
}
maxIndexes[i] = values[i].length - 1;
currentIndexes[i] = 0;
}
}
@Override
public boolean hasNext() {
return hasNext;
}
@Override
public String[] next() {
if (!hasNext) {
throw new NoSuchElementException("No more combinations are available");
}
final String[] combination = getCombinationByCurrentIndexes();
nextIndexesCombination();
return combination;
}
private String[] getCombinationByCurrentIndexes() {
final String[] combination = new String[combinationLength];
for (int i = 0; i < combinationLength; ++i) {
combination[i] = values[i][currentIndexes[i]];
}
return combination;
}
private void nextIndexesCombination() {
// A slightly modified "increment number by one" algorithm.
// This loop seems more natural, but it would return combinations in a different order than in your example:
// for (int i = 0; i < combinationLength; ++i) {
// This loop returns combinations in the order which matches your example:
for (int i = combinationLength - 1; i >= 0; --i) {
if (currentIndexes[i] < maxIndexes[i]) {
// Increment the current index
++currentIndexes[i];
return;
} else {
// Current index at max:
// reset it to zero and "carry" to the next index
currentIndexes[i] = 0;
}
}
// If we are here, then all current indexes are at max, and there are no more combinations
hasNext = false;
}
@Override
public void remove() {
throw new UnsupportedOperationException("Remove operation is not supported");
}
}
此处的示例用法:
final Map<Integer,String[]> map = new HashMap<Integer,String[]>();
map.put(1, new String[]{"test1", "stackoverflow"});
map.put(2, new String[]{"test2", "wow"});
map.put(3, new String[]{"new"});
final CombinationsIterator iterator = new CombinationsIterator(map);
while (iterator.hasNext()) {
System.out.println(
org.apache.commons.lang3.ArrayUtils.toString(iterator.next())
);
}
它会准确打印您示例中指定的内容。
P.S。实际上不需要地图;它可以被一个简单的数组(或列表列表)替换。然后构造函数会变得更简单:
public CombinationsIterator(final String[][] array) {
combinationLength = array.length;
values = array;
// ...
// Reorganize the map to array - THIS CAN BE REMOVED.
答案 1 :(得分:1)
我认为这是一个挑战,看看新的Java 8 API是否有助于解决这些问题。所以这是我解决问题的方法:
public class CombinatorIterator implements Iterator<Collection<String>> {
private final String[][] arrays;
private final int[] indices;
private final int total;
private int counter;
public CombinatorIterator(Collection<String[]> input) {
arrays = input.toArray(new String[input.size()][]);
indices = new int[arrays.length];
total = Arrays.stream(arrays).mapToInt(arr -> arr.length)
.reduce((x, y) -> x * y).orElse(0);
counter = 0;
}
@Override
public boolean hasNext() {
return counter < total;
}
@Override
public Collection<String> next() {
List<String> nextValue = IntStream.range(0, arrays.length)
.mapToObj(i -> arrays[i][indices[i]]).collect(Collectors.toList());
//rolling carry over the indices
for (int j = 0;
j < arrays.length && ++indices[j] == arrays[j].length; j++) {
indices[j] = 0;
}
counter++;
return nextValue;
}
}
请注意,我不使用地图作为输入,因为地图键在这里实际上不起任何作用。您可以使用map.values()
来传递迭代器的输入。使用以下测试代码:
List<String[]> input = Arrays.asList(
new String[] {"such", "nice", "question"},
new String[] {"much", "iterator"},
new String[] {"very", "wow"}
);
Iterator<Collection<String>> it = new CombinatorIterator(input);
it.forEachRemaining(System.out::println);
输出将是:
[such, much, very]
[nice, much, very]
[question, much, very]
[such, iterator, very]
[nice, iterator, very]
[question, iterator, very]
[such, much, wow]
[nice, much, wow]
[question, much, wow]
[such, iterator, wow]
[nice, iterator, wow]
[question, iterator, wow]