作者:恐龙 | 来源:互联网 | 2023-02-13 18:08
我对hashsets并不是很好,我想知道我是否能够使用.addAll方法检索hashets删除的重复项.所以在我的代码中,它要求用户输入最多20的值,如果它是-1,则中断.然后删除重复项并显示完成的列表.但现在我想显示已删除重复项的列表列表.例如:2 = 3个重复5 = 5个重复.这是我的代码:
import java.util.ArrayList;
import java.util.LinkedHashSet;
import java.util.Scanner;
import java.util.Set;
public class Question {
public static void main(String[] args) {
ArrayList entries = new ArrayList();
Scanner input = new Scanner(System.in);
int counter = 0;
int sentinel = -1;
while (entries.size() <= 20) {
System.out.println("Please enter some numbers into the entries list");
entries.add(input.nextInt());
if (entries.get(counter) == sentinel) {
entries.remove(counter);
break;
}
System.out.println("You've added: " + entries.get(counter) + ". So far your list is as shows: " + entries);
counter++;
}
System.out.println("The list is: " + entries);
System.out.println("Checking for Duplicates...");
Set num = new LinkedHashSet<>(entries);
num.addAll(entries); // The addAll method deletes duplicates
System.out.println(num);
entries.clear();
entries.addAll(num);
System.out.println("The Finished list is: " + entries);
}
}
CKing..
5
不幸的是,a Set
不会跟踪它拒绝的所有项目.您必须手动执行此操作.快速而肮脏的方法如下:
Set noDuplicates = new LinkedHashSet();
List rejects = new ArrayList();
for(Integer entry : entry) {
if(!noDuplicates.add(entry)) { rejects.add(entry) }
}
请注意,当尝试向a添加重复元素时,该add
方法返回false
并拒绝添加请求Set
.
1> CKing..:
不幸的是,a Set
不会跟踪它拒绝的所有项目.您必须手动执行此操作.快速而肮脏的方法如下:
Set noDuplicates = new LinkedHashSet();
List rejects = new ArrayList();
for(Integer entry : entry) {
if(!noDuplicates.add(entry)) { rejects.add(entry) }
}
请注意,当尝试向a添加重复元素时,该add
方法返回false
并拒绝添加请求Set
.