CollectionUtils方法

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.List;
import java.util.Map;

import org.apache.commons.collections.CollectionUtils;

public class LearnCollection {
public static void main(String[] args) {
List<Integer> num1=new ArrayList<Integer>();
num1.addAll(Arrays.asList(1,2,3,3));
System.out.println("num1:"+num1);

List<Integer> num2=new ArrayList<Integer>();
num2.addAll(Arrays.asList(3,3,3,4,5));
System.out.println("num2:"+num2);

/**
* union 并集
* The cardinality of each element in the returned Collection will be
* equal to the maximum of the cardinality of that element in the two
* given Collections.
*/
Collection<Integer> res1=CollectionUtils.union(num1, num2);
System.out.println("并集："+res1);

/**
* intersection 交集
* The cardinality of each element in the returned Collection will be
* equal to the minimum of the cardinality of that element in the two given
* Collections.
*
*/
Collection<Integer> res2=CollectionUtils.intersection(num1, num2);
System.out.println("交集："+res2);
/**
* 是否存在交集 containsAny
*/
System.out.println("是否存在交集："+CollectionUtils.containsAny(num1, num2));

/**
* 交集的补集 disjunction
* equivalent to subtract(union(a,b),intersection(a,b))
*/
Collection<Integer> res3=CollectionUtils.disjunction(num1, num2);
System.out.println("交集的补集："+res3);
/**
* 差集 subtract
* The cardinality of each element e in the returned Collection
* will be the cardinality of e in a minus the cardinality of e in b, or zero
*/
Collection<Integer> res4=CollectionUtils.subtract(num2, num1);
System.out.println("差集："+res4);
/**
* 判断是否是子集 集合A中每个元素的个数都大于另一个B中对应元素的个数，B是A的子集
*/
System.out.println("num1是不是num2的子集："+CollectionUtils.isSubCollection(num1, num2));


/**
* 还有很多其他函数，不一一列举
*/
/**
* 返回一个map，统计每个元素出现的次数
* Returns a Map mapping each unique element in the given Collection
* to an Integer representing the number of occurrences of that element
* in the Collection.
*/
Map<Integer, Integer> map=CollectionUtils.getCardinalityMap(num2);
System.out.println("元素--个数："+map);

}

}