java两个集合的交集

1. 用java编写程序，求集合的并集、交集和差集

publicstaticvoidmain(String[]args){
	Integer[]A={1,2,3,4};
	Integer[]B={1,3,7,9,11};
	
	List<Integer>listA=Arrays.asList(A);
	List<Integer>listB=Arrays.asList(B);
	
	List<Integer>jiaoji=newArrayList<Integer>();
	for(Integera:listA){
		if(listB.contains(a)){
			jiaoji.add(a);
		}
	}
	System.out.println(jiaoji);
	List<Integer>bingji=newArrayList<Integer>();
	for(Integera:listA){
		if(!bingji.contains(a)){
			bingji.add(a);
		}
	}
	for(Integerb:listB){
		if(!bingji.contains(b)){
			bingji.add(b);
		}
	}
	
	System.out.println(bingji);
	
	List<Integer>chaji=newArrayList<Integer>();
	for(Integera:listA){
		if(!listB.contains(a)){
			chaji.add(a);
		}
	}
	System.out.println(chaji);
}

2. java学习中，怎么求交集和并集

publicstaticvoidtest3(){
		Set<String>cc=newHashSet<String>();
		cc.add("2");
		cc.add("a");
		cc.add("b");
		cc.add("x");
		Set<String>dd=newHashSet<String>();
		dd.add("b");
		dd.add("1");
		dd.add("2");
		dd.add("3");
		dd.add("a");
		Set<String>ff=newHashSet<String>();
		ff.addAll(cc);
		ff.addAll(dd);
		System.out.println(ff.toString());
	}

这是并集

3. Java求两Collection的交集

intersection（）没问题。使用1.7测试代码如下：

importjava.util.ArrayList;
importjava.util.Collection;

publicclassTest{
	publicstaticvoidmain(String[]args){
		String[]arr1={"M116","M140","M250","M120","M98"};
		String[]arr2={"M116","M140","M250","M187","M98","M120"};
		Collection<String>collection1=toCollection(arr1);
		Collection<String>collection2=toCollection(arr2);
		Collection<String>collection=intersection(collection1,collection2);
		for(Strings:collection){
			System.out.println(s);
		}
	}

	publicstatic<T>Collection<T>intersection(Collection<T>collection1,
			Collection<T>collection2){
		Collection<T>collection=newArrayList<T>();//Howwoulditbe??
		for(Tt:collection1){
			if(collection2.contains(t)){
				collection.add(t);
			}
		}
		returncollection;
	}

	publicstatic<T>Collection<T>toCollection(T[]arr){
		Collection<T>collection=newArrayList<T>();
		for(Tt:arr){
			collection.add(t);
		}
		returncollection;
	}
}

4. JAVA怎么取多个List集合的交集

1、把多个list放到一个list中，生成 List<List<Object>>结构
2、遍历list，使用java8的规约操作，两两求交集
list.stream().rece((list1,list2) -> { list1.retainAll(list2); return list1;}).orElse(emptyList());

5. java找到两个list的交集并集差集

//橘散庆交集
set1.retainAll(set2);
//差圆握集
set1.removeAll(set2);
//并掘举集1

set1.addAll(set2);

6. java求交集和并集

处理集合交并集运算的类，实现java专门的集合类
简单实现代码如下：
package test;

import java.util.HashSet;
import java.util.Set;

public class Test {

public static void main(String[] args) {
Set<Integer> result = new HashSet<Integer>();
Set<Integer> set1 = new HashSet<Integer>(){{
add(1);
add(3);
add(5);
}};

Set<Integer> set2 = new HashSet<Integer>(){{
add(1);
add(2);
add(3);
}};

result.clear();
result.addAll(set1);
result.retainAll(set2);
System.out.println("交集："+result);

result.clear();
result.addAll(set1);
result.addAll(set2);
System.out.println("并集："+result);

}

}

7. java 怎么设计一个set类 然后new两个set对象 进行交集 并集 求代码

import java.util.Iterator;
import java.util.Set;
import java.util.TreeSet;

public class MySet {

public static void main(String[] args) {

Set set1 = new TreeSet();
set1.add("aa");
set1.add("bb");
set1.add("CC");

Set set2 = new TreeSet();
set2.add("aa");
set2.add("eeff");

Set unionSet = MySet.union(set1, set2);
Set intersectionSet = MySet.intersection(set1, set2);

printSet(unionSet);
System.out.println("-----------------------------");
printSet(intersectionSet);

}

private static void printSet(Set set) {

Iterator ite = set.iterator();

while(ite.hasNext()){
System.out.println(ite.next().toString());
}

}

public static Set union(Set set1, Set set2){
Set set = new TreeSet();

set.addAll(set1);
set.addAll(set2);

return set;
}

public static Set intersection(Set set1, Set set2){
Set set = new TreeSet();

if(set1.size() > set2.size()){
set.addAll(set2);
}else{
set.addAll(set1);
}

Iterator ite = set.iterator();

while(ite.hasNext()){
Object obj = ite.next();
if(!(set1.contains(obj) && set2.contains(obj))){
set.remove(obj);
}
}

return set;

}
}
--------------------------
CC
aa
bb
eeff
-----------------------------
aa