JAVA集合操作的利器:CollectionUtils
2015-07-16 18:05
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【IT168 技术】使用 CollectionUtils 中四个方法之一执行集合操作.这四种分别是 union(),intersection();disjunction();
subtract();
下列例子就是演示了如何使用上述四个方法处理两个 Collection;
注: 这些方法都是数学的集合算法
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.List;
import org.apache.commons.collections.CollectionUtils; import org.apache.commons.lang.ArrayUtils;
public class CollectionUtilsIntro {
@SuppressWarnings("unchecked")
public static void main(String[]
args) {
String[] arrayA = new String[] { "1", "2", "3", "3", "4", "5" };
String[] arrayB = new String[] { "3", "4", "4", "5", "6", "7" };
List<String> a = Arrays.asList(arrayA);
List<String> b = Arrays.asList(arrayB);
//并集
Collection<String> union = CollectionUtils.union(a,
b);
//交集
Collection<String> intersection = CollectionUtils.intersection(a,
b);
//交集的补集
Collection<String> disjunction = CollectionUtils.disjunction(a,
b);
//集合相减
Collection<String> subtract = CollectionUtils.subtract(a,
b);
Collections.sort((List<String>) union);
Collections.sort((List<String>) intersection);
Collections.sort((List<String>) disjunction);
Collections.sort((List<String>) subtract);
System.out.println("A: " + ArrayUtils.toString(a.toArray()));
System.out.println("B: " + ArrayUtils.toString(b.toArray()));
System.out.println("--------------------------------------------");
System.out.println("Union(A, B): " + ArrayUtils.toString(union.toArray()));
System.out.println("Intersection(A, B): " + ArrayUtils.toString(intersection.toArray()));
System.out.println("Disjunction(A, B): " + ArrayUtils.toString(disjunction.toArray()));
System.out.println("Subtract(A, B): " + ArrayUtils.toString(subtract.toArray()));
}
}
输出如下:
A: {1,2,3,3,4,5}
B: {3,4,4,5,6,7}
--------------------------------------------
Union(A, B): {1,2,3,3,4,4,5,6,7}
Intersection(A, B): {3,4,5}
Disjunction(A, B): {1,2,3,4,6,7}
<
b131
p style="color:rgb(51,51,51);font-family:Arial;font-size:14px;line-height:26px;">
Subtract(A, B): {1,2,3}
subtract();
下列例子就是演示了如何使用上述四个方法处理两个 Collection;
注: 这些方法都是数学的集合算法
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.List;
import org.apache.commons.collections.CollectionUtils; import org.apache.commons.lang.ArrayUtils;
public class CollectionUtilsIntro {
@SuppressWarnings("unchecked")
public static void main(String[]
args) {
String[] arrayA = new String[] { "1", "2", "3", "3", "4", "5" };
String[] arrayB = new String[] { "3", "4", "4", "5", "6", "7" };
List<String> a = Arrays.asList(arrayA);
List<String> b = Arrays.asList(arrayB);
//并集
Collection<String> union = CollectionUtils.union(a,
b);
//交集
Collection<String> intersection = CollectionUtils.intersection(a,
b);
//交集的补集
Collection<String> disjunction = CollectionUtils.disjunction(a,
b);
//集合相减
Collection<String> subtract = CollectionUtils.subtract(a,
b);
Collections.sort((List<String>) union);
Collections.sort((List<String>) intersection);
Collections.sort((List<String>) disjunction);
Collections.sort((List<String>) subtract);
System.out.println("A: " + ArrayUtils.toString(a.toArray()));
System.out.println("B: " + ArrayUtils.toString(b.toArray()));
System.out.println("--------------------------------------------");
System.out.println("Union(A, B): " + ArrayUtils.toString(union.toArray()));
System.out.println("Intersection(A, B): " + ArrayUtils.toString(intersection.toArray()));
System.out.println("Disjunction(A, B): " + ArrayUtils.toString(disjunction.toArray()));
System.out.println("Subtract(A, B): " + ArrayUtils.toString(subtract.toArray()));
}
}
输出如下:
A: {1,2,3,3,4,5}
B: {3,4,4,5,6,7}
--------------------------------------------
Union(A, B): {1,2,3,3,4,4,5,6,7}
Intersection(A, B): {3,4,5}
Disjunction(A, B): {1,2,3,4,6,7}
<
b131
p style="color:rgb(51,51,51);font-family:Arial;font-size:14px;line-height:26px;">
Subtract(A, B): {1,2,3}
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