这些对我来说就像集合操作：联合、重叠和差异。
看看这个：
Classical set operations for java.util.Collection
非常适合我。代码如下：
import java.util.ArrayList;
import java.util.List;
/**
* Add something descriptive here.
* User: mduffy
* Date: 3/26/2015
* Time: 1:27 PM
* @link https://stackoverflow.com/questions/29284061/find-the-uncommon-common-all-elements-from-two-different-array-list-objects-in/29284162?noredirect=1#comment46767251_29284162
*/
public class SetOperationDemo {
public static void main(String[] args) {
List<String> setOne = new ArrayList<String>() {{
add("A");
add("B");
add("C");
add("D");
add("E");
}};
List<String> setTwo = new ArrayList<String>() {{
add("D");
add("E");
add("F");
add("G");
}};
System.out.println("Set A : " + setOne);
System.out.println("Set B : " + setTwo);
List<String> base = new ArrayList<String>(setOne);
base.retainAll(setTwo);
System.out.println("Intersection A+B: " + base);
base = new ArrayList<String>(setOne);
base.removeAll(setTwo);
System.out.println("Subtraction A-B: " + base);
base = new ArrayList<String>(setTwo);
base.removeAll(setOne);
System.out.println("Subtraction B-A: " + base);
base = new ArrayList<String>(setOne);
base.addAll(setTwo);
System.out.println("Union A union B : " + base);
}
}
这是输出：
Set A : [A, B, C, D, E]
Set B : [D, E, F, G]
Intersection A+B: [D, E]
Subtraction A-B: [A, B, C]
Subtraction B-A: [F, G]
Union A union B : [A, B, C, D, E, D, E, F, G]
Process finished with exit code 0
如果您的列表包含自定义类，您必须确保它们正确地覆盖 equals 和 hashCode，否则它们不会给出预期的行为。这是我使用自定义类的代码，展示了它是如何完成的。
import java.util.ArrayList;
import java.util.List;
/**
* Add something descriptive here.
* User: mduffy
* Date: 3/26/2015
* Time: 1:27 PM
* @link https://stackoverflow.com/questions/29284061/find-the-uncommon-common-all-elements-from-two-different-array-list-objects-in/29284162?noredirect=1#comment46767251_29284162
*/
public class SetOperationDemo {
public static void main(String[] args) {
List<DemoPerson> setOne = new ArrayList<DemoPerson>() {{
add(new DemoPerson("Andy", "A"));
add(new DemoPerson("Bob", "B"));
add(new DemoPerson("Carl", "C"));
add(new DemoPerson("David", "D"));
add(new DemoPerson("Ernie", "E"));
}};
List<DemoPerson> setTwo = new ArrayList<DemoPerson>() {{
add(new DemoPerson("David", "D"));
add(new DemoPerson("Ernie", "E"));
add(new DemoPerson("Frank", "F"));
add(new DemoPerson("Gary", "G"));
}};
System.out.println("Set A : " + setOne);
System.out.println("Set B : " + setTwo);
List<DemoPerson> base = new ArrayList<DemoPerson>(setOne);
base.retainAll(setTwo);
System.out.println("Intersection A+B: " + base);
base = new ArrayList<DemoPerson>(setOne);
base.removeAll(setTwo);
System.out.println("Subtraction A-B: " + base);
base = new ArrayList<DemoPerson>(setTwo);
base.removeAll(setOne);
System.out.println("Subtraction B-A: " + base);
base = new ArrayList<DemoPerson>(setOne);
base.addAll(setTwo);
System.out.println("Union A union B : " + base);
}
}
class DemoPerson {
private final String firstName;
private final String lastName;
public DemoPerson(String firstName, String lastName) {
this.firstName = firstName;
this.lastName = lastName;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
DemoPerson that = (DemoPerson) o;
return !(firstName != null ? !firstName.equals(that.firstName) : that.firstName != null) && !(lastName != null ? !lastName.equals(that.lastName) : that.lastName != null);
}
@Override
public int hashCode() {
int result = firstName != null ? firstName.hashCode() : 0;
result = 31 * result + (lastName != null ? lastName.hashCode() : 0);
return result;
}
@Override
public String toString() {
final StringBuilder sb = new StringBuilder("{");
sb.append("'").append(firstName).append('\'');
sb.append(" '").append(lastName).append('\'');
sb.append('}');
return sb.toString();
}
}
这是输出，仍然按应有的方式运行：
Set A : [{'Andy' 'A'}, {'Bob' 'B'}, {'Carl' 'C'}, {'David' 'D'}, {'Ernie' 'E'}]
Set B : [{'David' 'D'}, {'Ernie' 'E'}, {'Frank' 'F'}, {'Gary' 'G'}]
Intersection A+B: [{'David' 'D'}, {'Ernie' 'E'}]
Subtraction A-B: [{'Andy' 'A'}, {'Bob' 'B'}, {'Carl' 'C'}]
Subtraction B-A: [{'Frank' 'F'}, {'Gary' 'G'}]
Union A union B : [{'Andy' 'A'}, {'Bob' 'B'}, {'Carl' 'C'}, {'David' 'D'}, {'Ernie' 'E'}, {'David' 'D'}, {'Ernie' 'E'}, {'Frank' 'F'}, {'Gary' 'G'}]
Process finished with exit code 0