使用 CollectionUtils 中四个方法之一执行集合操作.这四种分别是 union(),intersection();disjunction(); subtract();

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);
//并集
Collection<String> union = CollectionUtils.union(a, b);
//交集
Collection<String> intersection = CollectionUtils.intersection(a, b);
//交集的补集
Collection<String> disjunction = CollectionUtils.disjunction(a, b);
//集合相减
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()));
}
}

输出如下：

　　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}