下列例子就是演示瞭如何使用上述四個方法處理兩個 Collection;
注: 這些方法都是數學的集合算法
import java.util.Collection;
import java.util.Collections;
import java.util.List;
import org.apache.commons.collections.CollectionUtils; import org.apache.commons.lang.ArrayUtils;
public class CollectionUtilsIntro {
@SuppressWarnings("unchecked")
public static void main(String[] args) {
String[] arrayA = new String[] { "1", "2", "3", "3", "4", "5" };
String[] arrayB = new String[] { "3", "4", "4", "5", "6", "7" };
List<String> a = Arrays.asList(arrayA);
List<String> b = Arrays.asList(arrayB);
//並集
Collection<String> union = CollectionUtils.union(a, b);
//交集
Collection<String> intersection = CollectionUtils.intersection(a, b);
//交集的補集
Collection<String> disjunction = CollectionUtils.disjunction(a, b);
//集合相減
Collection<String> subtract = CollectionUtils.subtract(a, b);
Collections.sort((List<String>) union);
Collections.sort((List<String>) intersection);
Collections.sort((List<String>) disjunction);
Collections.sort((List<String>) subtract);
System.out.println("A: " + ArrayUtils.toString(a.toArray()));
System.out.println("B: " + ArrayUtils.toString(b.toArray()));
System.out.println("--------------------------------------------");
System.out.println("Union(A, B): " + ArrayUtils.toString(union.toArray()));
System.out.println("Intersection(A, B): " + ArrayUtils.toString(intersection.toArray()));
System.out.println("Disjunction(A, B): " + ArrayUtils.toString(disjunction.toArray()));
System.out.println("Subtract(A, B): " + ArrayUtils.toString(subtract.toArray()));
}
}
輸出如下:
A: {1,2,3,3,4,5}
B: {3,4,4,5,6,7}
--------------------------------------------
Union(A, B): {1,2,3,3,4,4,5,6,7}
Intersection(A, B): {3,4,5}
Disjunction(A, B): {1,2,3,4,6,7}
Subtract(A, B): {1,2,3}