Java判断两个集合是否具有交集以及如何获得交集 一、Java判断两个集合是否具有交集1、Collections.disjoint2、CollectionUtils.containsAny3、CollectionUtil.conta
全限定名为java.util.Collections,在俩个集合没有交集的时候会返回true,否则返回false。
比如:
List firstList = Arrays.asList("teacher", "worker", "student");List secondList = Arrays.asList("user", "admin");if (Collections.disjoint(firstList, secondList)) { System.out.println("Collections.disjoint方法:firstList、secondList没有交集");}
结果:
全限定名为:org.apache.commons.collections.CollectionUtils,两个集合有交集会返回true,否则会返回false,跟Collections.disjoint相反。
例如:
List firstList = Arrays.asList("teacher", "worker", "student");List secondList = Arrays.asList("user", "admin");if (CollectionUtils.containsAny(firstList, secondList)) { System.out.println("CollectionUtils.containsAny方法:firstList、secondList有交集");} else if (!CollectionUtils.containsAny(firstList, secondList)) { System.out.println("CollectionUtils.containsAny方法:firstList、secondList没有交集");}
结果:
全限定名为:cn.hutool.core.collection.CollectionUtil,两个集合有交集会返回true,否则会返回false,跟Collections.disjoint相反。
比如:
List firstList = Arrays.asList("teacher", "worker", "student"); List secondList = Arrays.asList("user", "admin"); if (CollectionUtil.containsAny(firstList, secondList)) { System.out.println("CollectionUtil.containsAny方法:firstList、secondList有交集"); } else if (!CollectionUtil.containsAny(firstList, secondList)) { System.out.println("CollectionUtil.containsAny方法:firstList、secondList没有交集"); }
结果:
Java8中有两种方式可以得到两个集合是否有交集,如下:
List firstList = Arrays.asList("teacher", "worker", "student");List secondList = Arrays.asList("user", "admin");//方式一//List resultList = firstList.stream().filter((firstItem) -> secondList.contains(firstItem)).collect(Collectors.toList());//方式二List resultList = firstList.stream().filter(secondList::contains).collect(Collectors.toList());if (resultList != null && resultList.size() > 0) { System.out.println("firstList、secondList有交集");} else { System.out.println("firstList、secondList没有交集");}
结果:
比如:
public static void main(String[] args) { List firstList = Arrays.asList("teacher", "worker", "student", "driver"); List secondList = Arrays.asList("user", "admin", "student", "driver"); List resultList = new ArrayList<>(); for (String item : firstList) { if (secondList.contains(item)) { resultList.add(item); } } System.out.println(resultList);}
结果:
要是两个集合中存在重复元素,这样得到的结果是没有去重的,如果是:
结果是:
如果要去重可以加上
最终的代码:
public static void main(String[] args) { List firstList = Arrays.asList("teacher", "worker", "student", "driver", "driver"); List secondList = Arrays.asList("user", "admin", "student", "driver", "driver"); List resultList = new ArrayList<>(); for (String item : firstList) { if (secondList.contains(item) && !resultList.contains(item)) { resultList.add(item); } } System.out.println(resultList);}
这样得到的结果就是去重的了
也可以使用HashSet去重
结果:
比如:
public static void main(String[] args) { List firstList = Arrays.asList("teacher", "worker", "student", "driver"); List secondList = Arrays.asList("user", "admin", "student", "driver"); List resultList = new ArrayList<>(); if (firstList == null) { throw new RuntimeException("firstList为空!"); } firstList.forEach((firstItem) -> { if (secondList.contains(firstItem)) { resultList.add(firstItem); } }); System.out.println(resultList);}
结果:
假如两个集合有相同的元素
和“二.1”一样,这样也是不能去重的
做法也是和上面那样
结果:
或者用HashSet,也和“二.1”一样。
和“一.4”那样,有两种方式可以得到两集合的交集。
代码如下:
public static void main(String[] args) { List firstList = Arrays.asList("teacher", "worker", "student"); List secondList = Arrays.asList("user", "admin", "student"); //方式一 List resultList = firstList.stream().filter((firstItem) -> secondList.contains(firstItem)).collect(Collectors.toList()); //方式二 //List resultList = firstList.stream().filter(secondList::contains).collect(Collectors.toList()); System.out.println(resultList);}
结果:
和“二.1”、“二.2”那样也是不能去除重复元素的
结果:
加上distinct就可以去重了
结果:
来源地址:https://blog.csdn.net/studio_1/article/details/129873235
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