取得集合list的交集并集（javaAPI或者apache的API）

javaAPI方式----------->

/** * 求ls对ls2的差集,即ls中有，但ls2中没有的 * * @param ls * @param ls2 * @return */public static List diff(List ls, List ls2) {    List list = new ArrayList(Arrays.asList(new Object[ls.size()]));    Collections.copy(list, ls);    list.removeAll(ls2);    return list;}/** * 求2个集合的交集 * * @param ls * @param ls2 * @return */public static List intersect(List ls, List ls2) {    List list = new ArrayList(Arrays.asList(new Object[ls.size()]));    Collections.copy(list, ls);    list.retainAll(ls2);    return list;}/** * 求2个集合的并集 * * @param ls * @param ls2 * @return */public static List union(List ls, List ls2) {    List list = new ArrayList(Arrays.asList(new Object[ls.size()]));    Collections.copy(list, ls);//将ls的值拷贝一份到list中    list.removeAll(ls2);    list.addAll(ls2);    return list;}

apacheAPI方式---------->

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();下列例子就是演示了如何使用上述四个方法处理两个 Collection;注: 这些方法都是数学的集合算法
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import java.util.Arrays;   
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);   
  //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}