Question

我有两个ArrayList<Integer>如下：

原文：12, 16, 17, 19, 101

已选中：16, 19, 107, 108, 109

我想在这些列表上做交集/联合，最后我有两个列表：

添加：108,109,107

删除：12, 17, 101

原始列表和所选列表的长度各不相同，一个可以比另一个更大/更小

Answer 1

作为替代方案，您可以使用Apache commons库中的CollectionUtils。它具有适合您案例的静态交集，联合和减去方法。

Answer 2

List<Integer> original = Arrays.asList(12,16,17,19,101);
List<Integer> selected = Arrays.asList(16,19,107,108,109);

ArrayList<Integer> add = new ArrayList<Integer>(selected);
add.removeAll(original);
System.out.println("Add: " + add);

ArrayList<Integer> remove = new ArrayList<Integer>(original);
remove.removeAll(selected);
System.out.println("Remove: " + remove);

输出：

Add: [107, 108, 109]
Remove: [12, 17, 101]

使用Collection的removeAll方法。 See javadocs.

Answer 3

交叉点：original.retainAll(selected)。

之后，原始文件将仅包含两个集合中的元素。如果有任何改变，则返回true。

警告：此方法对于大型集合来说非常慢

Answer 4

对于交集和联合操作，自然集合类型是Set而不是List，它的使用效率也更高。

Answer 5

使用Guava库：

List<Integer> listA = Lists.newArrayList(12,16,17,19,101);
List<Integer> listB = Lists.newArrayList(16,19,107,108,109);
Set<Integer> intersection = Sets.intersection(Sets.newHashSet(listA), Sets.newHashSet(listB));
listA.removeAll(intersection);
listB.removeAll(intersection);

Answer 6

 List<Integer> original;
 List<Integer> selected;

 List<Integer> add = new ArrayList<Integer>(selected);
 add.removeAll(original);

 List<Integer> remove = new ArrayList<Integer>(original);
 remove.removeAll(selected);

小心重复元素周围的边框情况。基数应该得到尊重吗？如果我原来5, 6和5, 5, 6之后，应该添加5？以上内容适用于Set，因为它们没有重复项（加contains()次查找更快，因为它们被包含的数据编入索引）。

Answer 7

如果要获取列表列表的交集，请使用此方法

List<Address> resultsIntersectionSet( List<Set<Address>> inputListOfLists )
{
    Set<Address> intersection = new HashSet<>();

    if ( !inputListOfLists.isEmpty() )
        intersection = inputListOfLists.get( 0 );

    for ( Set<Address> filterResultList : inputListOfLists )
    {
        intersection.retainAll( filterResultList );
    }

    return new ArrayList<>( intersection );
}

Answer 8

有一个新库underscore-java。它可以为列表和数组做差异和交集。 Live example.

代码示例：

FactTable.DateKey=BridgeTable.PYDateKey

Answer 9

package LAB8Pack;
import java.util.HashSet; 
import java.util.Iterator;
import java.lang.StringBuilder; 


public class HashSetDemo {

    public static void main(String[] args) {
        HashSet<String> round = new HashSet<String> (); 
        HashSet<String> green = new HashSet<String> (); 


        // Add elements to 'round' and 'green' sets 
        //----------------------------------------------------
        round.add("peas"); 
        green.add("peas");
        round.add("watermelon"); 
        green.add("watermelon");
        round.add("basketball"); 
        green.add("chameleon");
        round.add("chameleon"); 
        green.add("grass");
        round.add("eyes"); 
        green.add("book");

        // Create 'setUnion' and 'setInter'  
        // ---------------------------------------------------
        HashSet<String> setUnion = new HashSet<String>();   

        // Use this to find the intersection
        HashSet<String> SETINTER = new HashSet<String>();   


        HashSet<String> setInter1 = new HashSet<String>(round);
        HashSet<String> setInter2 = new HashSet<String>(green);



        // Add all the elements to one set
        setUnion.addAll(round);
        setUnion.addAll(green);
        SETINTER.addAll(setUnion);



        // Create an intersection
        setInter1.removeAll(green); // Get unique items in round
        setInter2.removeAll(round); // Get unique items in green
        SETINTER.removeAll(setInter2); // Remove items that are unique to green
        SETINTER.removeAll(setInter1); // Remove items that are unique to round
        //----------------------------------------------------


        // DISPLAY RESULTS 
        // ===================================================
        System.out.println("Content of set round"); 
        System.out.println("-----------------------");
        System.out.println(OutputSet(round));

        System.out.println("Content of set green");
        System.out.println("-----------------------");
        System.out.println(OutputSet(green)); 

        System.out.println("Content of set Union");
        System.out.println("-----------------------");
        System.out.println(OutputSet(setUnion));



        System.out.println("Content of set Intersection"); 
        System.out.println("-----------------------");
        System.out.println(OutputSet(SETINTER));
    }




    // METHODS 
    // =======================================================
    static StringBuilder OutputSet (HashSet<String> args) {
        Iterator iterator = args.iterator();
        StringBuilder sB = new StringBuilder (); 

        while (iterator.hasNext()) {
            sB.append(iterator.next() + " \n");
        }
        return sB; 
    }  
}

Answer 10

这里是查找各种集合（超过2个）的交集的功能-

public static <T, C extends Collection<T>> C findIntersection(C newCollection,
                                                            Collection<T>... collections) {
  boolean first = true;
  for (Collection<T> collection : collections) {
      if (first) {
          newCollection.addAll(collection);
          first = false;
      } else {
          newCollection.retainAll(collection);
      }
  }
  return newCollection;
}

用法-

public static void main(String[] args) {
  List<Integer> l1 = List.of(1, 3, 5, 7, 9, 11, 13);
  List<Integer> l2 = List.of(1, 2, 3, 5, 8, 13);
  List<Integer> l3 = List.of(2, 3, 5, 7, 11, 13);
  Set<Integer> intersection = findIntersection(new HashSet<>(), l1, l2, l3);
  System.out.println(intersection);
 }

Answer 11

使用Java 8+流：

List<Integer> result = original.stream()
  .distinct()
  .filter(selected::contains)
  .collect(Collectors.toList());

在java中列出交集

11 个答案: