我有以下 map :Map<Integer,String[]> map = new HashMap<Integer,String[]>();
键是整数,值是数组(也可以用列表代替)。
现在,我想获得键值的所有可能组合。例如,假设 map 包含以下条目:
key 1: "test1", "stackoverflow"
key 2: "test2", "wow"
key 3: "new"
组合包括
("test1","test2","new")
("test1","wow","new")
("stackoverflow", "test2", "new")
("stackoverflow", "wow", "new")
为此我设想了一个方法 boolean hasNext()
如果存在下一对,则返回 true,而第二个方法仅返回下一组值(如果有)。
如何做到这一点? map 也可以替换为其他数据结构。
最佳答案
该算法本质上与十进制数的递增算法(“x -> 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");
}
}
这里是示例用法:
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())
);
}
它会准确打印您的示例中指定的内容。
附言实际上不需要 map ;它可以由一个简单的数组数组(或列表列表)代替。构造函数会变得更简单一些:
public CombinationsIterator(final String[][] array) {
combinationLength = array.length;
values = array;
// ...
// Reorganize the map to array - THIS CAN BE REMOVED.
关于java - 如何在 Java 中创建值的组合?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35961157/