鑑於您已經說過您的兩個列表已經排序,它們可以在O(N)時間內進行比較,這比您使用ListUtils的當前解決方案快得多。以下方法使用類似算法來合併兩個排序列表,這些列表可以在大多數教科書中找到。
import java.util.*;
public class CompareSortedLists {
public static void main(String[] args) {
List<Integer> sourceDbResults = Arrays.asList(1, 2, 3, 4, 5, 8);
List<Integer> hiveResults = Arrays.asList(2, 3, 6, 7);
List<Integer> inSourceDb_notInHive = new ArrayList<>();
List<Integer> inHive_notInSourceDb = new ArrayList<>();
compareSortedLists(
sourceDbResults, hiveResults,
inSourceDb_notInHive, inHive_notInSourceDb);
assert inSourceDb_notInHive.equals(Arrays.asList(1, 4, 5, 8));
assert inHive_notInSourceDb.equals(Arrays.asList(6, 7));
}
/**
* Compares two sorted lists (or other iterable collections in ascending order).
* Adds to onlyInList1 any and all elements in list1 that are not in list2; and
* conversely to onlyInList2. The caller must ensure the two input lists are
* already sorted and should initialize onlyInList1 and onlyInList2 to empty,
* writable collections.
*/
public static <T extends Comparable<? super T>> void compareSortedLists(
Iterable<T> list1, Iterable<T> list2,
Collection<T> onlyInList1, Collection<T> onlyInList2) {
Iterator<T> it1 = list1.iterator();
Iterator<T> it2 = list2.iterator();
T e1 = it1.hasNext() ? it1.next() : null;
T e2 = it2.hasNext() ? it2.next() : null;
while (e1 != null || e2 != null) {
if (e2 == null) { // No more elements in list2, some remaining in list1
onlyInList1.add(e1);
e1 = it1.hasNext() ? it1.next() : null;
}
else if (e1 == null) { // No more elements in list1, some remaining in list2
onlyInList2.add(e2);
e2 = it2.hasNext() ? it2.next() : null;
}
else {
int comp = e1.compareTo(e2);
if (comp < 0) {
onlyInList1.add(e1);
e1 = it1.hasNext() ? it1.next() : null;
}
else if (comp > 0) {
onlyInList2.add(e2);
e2 = it2.hasNext() ? it2.next() : null;
}
else /* comp == 0 */ {
e1 = it1.hasNext() ? it1.next() : null;
e2 = it2.hasNext() ? it2.next() : null;
}
}
}
}
}
上述方法不使用外部庫,可以使用任何版本的Java,從6開始。如果使用PeekingIterator,比如Apache Commons Collections中,或番石榴的人,或者自己寫,那麼你就可以使該方法更簡單,特別是如果你還使用Java 8:
public static <T extends Comparable<? super T>> void compareSortedLists(
Iterable<T> list1, Iterable<T> list2,
Collection<T> onlyInList1, Collection<T> onlyInList2) {
PeekingIterator<T> it1 = new PeekingIterator<>(list1.iterator());
PeekingIterator<T> it2 = new PeekingIterator<>(list2.iterator());
while (it1.hasNext() && it2.hasNext()) {
int comp = it1.peek().compareTo(it2.peek());
if (comp < 0)
onlyInList1.add(it1.next());
else if (comp > 0)
onlyInList2.add(it2.next());
else /* comp == 0 */ {
it1.next();
it2.next();
}
}
it1.forEachRemaining(onlyInList1::add);
it2.forEachRemaining(onlyInList2::add);
}
重新開放。似乎不像http://stackoverflow.com/questions/41608074/comparing-2-very-large-arraylists-in-java的重複。在另外一個問題中,名單隻有10萬個,而且由於某些未知原因,問題已經耗盡。這個問題似乎更多關於算法。 –
你只需要知道2列表是否相等?元素的順序是重要的嗎?你是否需要其他信息,如list1是否是其他信息的子集? – 6ton
你能通過比較兩個列表來更好地描述你的意思嗎? –