设置交集（）函数|番石榴 |Java

Guava 的Sets.intersection()返回两个集合交集的不可修改视图。返回的集合包含两个支持集合所包含的所有元素。返回集合的迭代顺序与 set1 的迭代顺序一致。

句法：

public static  
    Sets.SetView 
        intersection(Set set1, Set set2)

返回值：此方法返回两个集合交集的不可修改视图。

以下示例说明了 Sets 交集方法的工作原理：

示例 1：

// Java code to show implementation
// of Guava's Sets.intersection() method
  
import com.google.common.collect.Sets;
import java.util.Set;
  
class GFG {
  
    // Driver's code
    public static void main(String[] args)
    {
        // Creating first set
        Set
            set1 = Sets.newHashSet(10, 20, 30, 40, 50);
  
        // Creating second set
        Set
            set2 = Sets.newHashSet(30, 50, 70, 90);
  
        // Using Guava's Sets.intersection() method
        Set
            answer = Sets.intersection(set1, set2);
  
        // Displaying the intersection of set1 and set2
        System.out.println("Set 1: "
                           + set1);
        System.out.println("Set 2: "
                           + set2);
        System.out.println("Set 1 intersection Set 2: "
                           + answer);
    }
}

输出：

Set 1: [40, 10, 50, 20, 30]
Set 2: [50, 90, 30, 70]
Set 1 intersection Set 2: [50, 30]

示例 2：

// Java code to show implementation
// of Guava's Sets.intersection() method
  
import com.google.common.collect.Sets;
import java.util.Set;
  
class GFG {
  
    // Driver's code
    public static void main(String[] args)
    {
        // Creating first set
        Set
            set1 = Sets.newHashSet("G", "e", "e", "k", "s");
  
        // Creating second set
        Set
            set2 = Sets.newHashSet("g", "f", "G", "e");
  
        // Using Guava's Sets.intersection() method
        Set
            answer = Sets.intersection(set1, set2);
  
        // Displaying the intersection of set1 and set2
        System.out.println("Set 1: "
                           + set1);
        System.out.println("Set 2: "
                           + set2);
        System.out.println("Set 1 intersection Set 2: "
                           + answer);
    }
}

输出：

Set 1: [k, s, e, G]
Set 2: [e, f, g, G]
Set 1 intersection Set 2: [e, G]

注意：当 set1 是两个集合中较小的一个时，返回的视图性能稍好。如果您有理由相信您的一组通常会比另一组小，请先通过它。