条件:不修改原有列表;只使用JDK,没有外部库。单行程序或JDK 1.3版本的加分项。
有没有比这更简单的方法:
List<String> newList = new ArrayList<String>();
newList.addAll(listOne);
newList.addAll(listTwo);
条件:不修改原有列表;只使用JDK,没有外部库。单行程序或JDK 1.3版本的加分项。
有没有比这更简单的方法:
List<String> newList = new ArrayList<String>();
newList.addAll(listOne);
newList.addAll(listTwo);
当前回答
不是更简单,但没有调整开销:
List<String> newList = new ArrayList<>(listOne.size() + listTwo.size());
newList.addAll(listOne);
newList.addAll(listTwo);
其他回答
在一般情况下,如果不介绍您自己的实用程序方法,我无法改进双行程序,但如果您确实有字符串列表,并且您愿意假设这些字符串不包含逗号,您可以使用这个长一行程序:
List<String> newList = new ArrayList<String>(Arrays.asList((listOne.toString().subString(1, listOne.length() - 1) + ", " + listTwo.toString().subString(1, listTwo.length() - 1)).split(", ")));
如果去掉泛型,这应该是JDK 1.4兼容的(尽管我还没有测试)。也不建议用于生产代码;-)
没有办法接近一行程序,但我认为这是最简单的:
List<String> newList = new ArrayList<String>(l1);
newList.addAll(l2);
for(String w:newList)
System.out.printf("%s ", w);
Java 8版本,支持通过对象键连接:
public List<SomeClass> mergeLists(final List<SomeClass> left, final List<SomeClass> right, String primaryKey) {
final Map<Object, SomeClass> mergedList = new LinkedHashMap<>();
Stream.concat(left.stream(), right.stream())
.map(someObject -> new Pair<Object, SomeClass>(someObject.getSomeKey(), someObject))
.forEach(pair-> mergedList.put(pair.getKey(), pair.getValue()));
return new ArrayList<>(mergedList.values());
}
我们可以用两种方法使用java8连接两个列表。
List<String> list1 = Arrays.asList("S", "T");
List<String> list2 = Arrays.asList("U", "V");
1)使用concat:
List<String> collect2 = Stream.concat(list1.stream(), list2.stream()).collect(toList());
System.out.println("collect2 = " + collect2); // collect2 = [S, T, U, V]
2)使用flatMap:
List<String> collect3 = Stream.of(list1, list2).flatMap(Collection::stream).collect(toList());
System.out.println("collect3 = " + collect3); // collect3 = [S, T, U, V]
import java.util.AbstractList;
import java.util.List;
/**
* The {@code ConcatList} is a lightweight view of two {@code List}s.
* <p>
* This implementation is <em>not</em> thread-safe even though the underlying lists can be.
*
* @param <E>
* the type of elements in this list
*/
public class ConcatList<E> extends AbstractList<E> {
/** The first underlying list. */
private final List<E> list1;
/** The second underlying list. */
private final List<E> list2;
/**
* Constructs a new {@code ConcatList} from the given two lists.
*
* @param list1
* the first list
* @param list2
* the second list
*/
public ConcatList(final List<E> list1, final List<E> list2) {
this.list1 = list1;
this.list2 = list2;
}
@Override
public E get(final int index) {
return getList(index).get(getListIndex(index));
}
@Override
public E set(final int index, final E element) {
return getList(index).set(getListIndex(index), element);
}
@Override
public void add(final int index, final E element) {
getList(index).add(getListIndex(index), element);
}
@Override
public E remove(final int index) {
return getList(index).remove(getListIndex(index));
}
@Override
public int size() {
return list1.size() + list2.size();
}
@Override
public boolean contains(final Object o) {
return list1.contains(o) || list2.contains(o);
}
@Override
public void clear() {
list1.clear();
list2.clear();
}
/**
* Returns the index within the corresponding list related to the given index.
*
* @param index
* the index in this list
*
* @return the index of the underlying list
*/
private int getListIndex(final int index) {
final int size1 = list1.size();
return index >= size1 ? index - size1 : index;
}
/**
* Returns the list that corresponds to the given index.
*
* @param index
* the index in this list
*
* @return the underlying list that corresponds to that index
*/
private List<E> getList(final int index) {
return index >= list1.size() ? list2 : list1;
}
}