Java judges whether two sets have intersection and how to get the intersection

1. Java judges whether two sets have intersection

1、Collections.disjoint

The fully qualified name is java.util.Collections, and it will return true when the two collections do not intersect, otherwise it will return false.

for example:

List<String> firstList = Arrays.asList("teacher", "worker", "student");
List<String> secondList = Arrays.asList("user", "admin");

if (Collections.disjoint(firstList, secondList)) {
    
    
    System.out.println("Collections.disjoint方法：firstList、secondList没有交集");
}

result:

2、CollectionUtils.containsAny

The fully qualified name is: org.apache.commons.collections.CollectionUtils. If two collections intersect, it will return true, otherwise it will return false, which is the opposite of Collections.disjoint.

For example:

List<String> firstList = Arrays.asList("teacher", "worker", "student");
List<String> 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");
}

result:

3、CollectionUtil.containsAny

The fully qualified name is: cn.hutool.core.collection.CollectionUtil. If two collections intersect, it will return true, otherwise it will return false, which is the opposite of Collections.disjoint.

for example:

 List<String> firstList = Arrays.asList("teacher", "worker", "student");
 List<String> 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");
 }

result:

4. Use the new features of Java8

There are two ways in Java8 to get whether two collections intersect, as follows:

List<String> firstList = Arrays.asList("teacher", "worker", "student");
List<String> secondList = Arrays.asList("user", "admin");

//方式一
//List<String> resultList = firstList.stream().filter((firstItem) -> secondList.contains(firstItem)).collect(Collectors.toList());

//方式二
List<String> 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");
}

result:

2. Obtain the intersection of two sets

1. Use for loop

for example:

public static void main(String[] args) {
    
    
    List<String> firstList = Arrays.asList("teacher", "worker", "student", "driver");
    List<String> secondList = Arrays.asList("user", "admin", "student", "driver");

    List<String> resultList = new ArrayList<>();
    for (String item : firstList) {
    
    
        if (secondList.contains(item)) {
    
    
            resultList.add(item);
        }
    }
    System.out.println(resultList);
}

result:

If there are duplicate elements in the two collections, the result obtained in this way is not deduplicated, if it is:

turn out:

If you want to add weight, you can add

The final code:

public static void main(String[] args) {
    
    
    List<String> firstList = Arrays.asList("teacher", "worker", "student", "driver", "driver");
    List<String> secondList = Arrays.asList("user", "admin", "student", "driver", "driver");

    List<String> resultList = new ArrayList<>();
    for (String item : firstList) {
    
    
        if (secondList.contains(item) && !resultList.contains(item)) {
    
    
            resultList.add(item);
        }
    }
    System.out.println(resultList);
}

The result obtained in this way is deduplication.

You can also use HashSet to deduplicate

result:

2. Use Java8's forEach

for example:

public static void main(String[] args) {
    
    
    List<String> firstList = Arrays.asList("teacher", "worker", "student", "driver");
    List<String> secondList = Arrays.asList("user", "admin", "student", "driver");

    List<String> 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);
}

result:

If two sets have the same elements

Like "2.1", it cannot be deduplicated

The method is also the same as above

result:

Or use HashSet, which is the same as "2.1".

3. Use the new features of Java8

Like "1.4", there are two ways to get the intersection of two sets.

code show as below:

public static void main(String[] args) {
    
    
    List<String> firstList = Arrays.asList("teacher", "worker", "student");
    List<String> secondList = Arrays.asList("user", "admin", "student");

    //方式一
    List<String> resultList = firstList.stream().filter((firstItem) -> secondList.contains(firstItem)).collect(Collectors.toList());

    //方式二
    //List<String> resultList = firstList.stream().filter(secondList::contains).collect(Collectors.toList());
    System.out.println(resultList);
}

result:

Like "2.1" and "2.2", it is also impossible to remove repeated elements

result:

Adding distinct can deduplicate

the result: