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  • CollectionUtils常用方法

    CollectionUtils常用方法

    • 是否存在交集:boolean isContained = CollectionUtils.containsAny(aList, bList);
    • 并集:Collection<String> unionList = CollectionUtils.union(aList, bList);
    • 交集:Collection<String> intersectionList = CollectionUtils.intersection(aList, bList);
    • 交集的补集:Collection<String> disjunctionList = CollectionUtils.disjunction(aList, bList);
    • 集合相减:Collection<String> subtractList = CollectionUtils.subtract(aList, bList);

    依赖jar包:commons-collections4-4.4.jar

    参考API:http://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/CollectionUtils.html

    代码测试

    import java.util.Arrays;
    import java.util.Collection;
    import java.util.Collections;
    import java.util.List;
    import org.apache.commons.collections4.CollectionUtils;
    import org.apache.commons.lang.ArrayUtils;
    
    /**
     * @author 踏步
     * @version 创建时间:2020年5月27日下午2:00:00
     * @Description:TODO 测试类
     */
    public class TestCollectionUtils {
        public static void main(String[] args) {
            String[] arrayA = new String[] { "A", "B", "C", "D", "E", "F" };  
            String[] arrayB = new String[] { "B", "D", "F", "G", "H", "K" };
            List<String> aList = Arrays.asList(arrayA);
            List<String> bList = Arrays.asList(arrayB);
            // 是否存在交集
            boolean isContained = CollectionUtils.containsAny(aList, bList);
            // 并集
            Collection<String> unionList = CollectionUtils.union(aList, bList);
            // 交集
            Collection<String> intersectionList = CollectionUtils.intersection(aList, bList);
            // 交集的补集
            Collection<String> disjunctionList = CollectionUtils.disjunction(aList, bList);
            // 集合相减
            Collection<String> subtractList = CollectionUtils.subtract(aList, bList);
            // 排序
            Collections.sort((List<String>) unionList);
            Collections.sort((List<String>) intersectionList);
            Collections.sort((List<String>) disjunctionList);
            Collections.sort((List<String>) subtractList);
            // 测试结果
            System.out.println("集合--A: " + ArrayUtils.toString(aList.toArray()));
            System.out.println("集合--B: " + ArrayUtils.toString(bList.toArray()));
            System.out.println("是否存在交集--A has one of B? : " + isContained);
            System.out.println("并集--Union(A, B): " + ArrayUtils.toString(unionList.toArray()));
            System.out.println("交集--Intersection(A, B): " + ArrayUtils.toString(intersectionList.toArray()));
            System.out.println("交集的补集--Disjunction(A, B): " + ArrayUtils.toString(disjunctionList.toArray()));
            System.out.println("集合相减--Subtract(A, B): " + ArrayUtils.toString(subtractList.toArray()));
        }
    }

     测试结果

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  • 原文地址:https://www.cnblogs.com/mjtabu/p/12973000.html
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