生成产品时采用延迟加载的方式，即仅在访问元组时才创建它，这种方法如何？
abstract public class CartesianProductView<T> extends AbstractList<T> {
private final List<List<T>> factors;
private final int size;
public CartesianProductView(List<List<T>> factors) {
this.factors = new ArrayList<>(factors);
Collections.reverse(this.factors);
int acc = 1;
for (Iterator<List<T>> iter = this.factors.iterator(); iter.hasNext(); ) {
acc *= iter.next().size();
}
this.size = acc;
}
@Override
public T get(int index) {
if (index < 0 || index >= size()) {
throw new IndexOutOfBoundsException(String.format("index %d > size() %d", index, size()));
}
T acc = null;
for (Iterator<List<T>> iter = factors.iterator(); iter.hasNext();) {
List<T> set = iter.next();
acc = makeTupleOrSingleton(set.get(index % set.size()), acc);
index /= set.size();
}
return acc;
}
@Override
public int size() {
return size;
}
private T makeTupleOrSingleton(T left, T right) {
if (right == null) {
return left;
}
return makeTuple(left, right);
}
abstract protected T makeTuple(T left, T right);
}
并像这样使用它
final List<List<String>> l1 = new ArrayList<List<String>>() {{ add(singletonList("a")); add(singletonList("b")); add(singletonList("c")); }};
final List<List<String>> l2 = new ArrayList<List<String>>() {{ add(singletonList("X")); add(singletonList("Y")); }};
final List<List<String>> l3 = new ArrayList<List<String>>() {{ add(singletonList("1")); add(singletonList("2")); add(singletonList("3")); add(singletonList("4")); }};
List<List<List<String>>> in = new ArrayList<List<List<String>>>() {{ add(l1); add(l2); add(l3); }};
List<List<String>> a = new CartesianProductView<List<String>>(in) {
@Override
protected List<String> makeTuple(final List<String> left, final List<String> right) {
return new ArrayList<String>() {{ add(left.get(0)); addAll(right); }};
}
};
System.out.println(a);
结果如下：
[[a, X, 1], [a, X, 2], [a, X, 3], [a, X, 4], [a, Y, 1], [a, Y, 2], [a, Y, 3], [a, Y, 4], [b, X, 1], [b, X, 2], [b, X, 3], [b, X, 4], [b, Y, 1], [b, Y, 2], [b, Y, 3], [b, Y, 4], [c, X, 1], [c, X, 2], [c, X, 3], [c, X, 4], [c, Y, 1], [c, Y, 2], [c, Y, 3], [c, Y, 4]]
作为额外的奖励，您可以使用它将所有字符串连接起来：
final List<String> l1 = new ArrayList<String>() {{ add("a"); add("b"); add("c"); }};
final List<String> l2 = new ArrayList<String>() {{ add("X"); add("Y"); }};
final List<String> l3 = new ArrayList<String>() {{ add("1"); add("2"); add("3"); add("4"); }};
List<List<String>> in = new ArrayList<List<String>>() {{ add(l1); add(l2); add(l3); }};
List<String> a = new CartesianProductView<String>(in) {
@Override
protected String makeTuple(String left, String right) {
return String.format("%s%s", left, right);
}
};
System.out.println(a);
结果如下：
[aX1, aX2, aX3, aX4, aY1, aY2, aY3, aY4, bX1, bX2, bX3, bX4, bY1, bY2, bY3, bY4, cX1, cX2, cX3, cX4, cY1, cY2, cY3, cY4]
Vector？https://dev59.com/WnM_5IYBdhLWcg3waieJ - josh.trowInteger键在LHM中的意义是什么？组合数组的排序应该基于键的顺序还是LHM的顺序？（例如，如果我按照3,2,1的顺序添加键到LHM中，那么我应该使用相同的排序还是1,2,3）假设我有键1,2,4,6的映射...组合数组应该跳过3和5或使用一些特殊值（例如0，-1等）？ - Kevin KString的例子更清楚了。 - Kevin K