使用remove()或remove()从ArrayList中删除项

Question

有人向我推荐了两种实现这一目标的方法：

这个'search'方法应该返回ArrayList中的位置（以便可以使用remove( ) )或对象引用（以便可以使用remove( ) )。如果找不到工作人员，则搜索方法应返回-1（如果使用remove( ) )或null（如果使用remove( ) )。

然而，我不确定如何在Java中实现这一点。

import java.util.*;

import personnelPackage.Personnel;

public class ArrayListTest
{
    static Scanner keyboard = new Scanner(System.in);

    public static void main(String[] args)
    {
        long searchQuery;

        ArrayList<Personnel> staffList = new ArrayList<Personnel>();
        Personnel[] staff =
            {new Personnel(123456,"Smith","John"),
             new Personnel(234567,"Jones","Sally Ann"),
             new Personnel(999999,"Black","James Paul")};

        for (Personnel person:staff)
            staffList.add(person);

        do
        {
            showDisplay(staffList);

            System.out.print("\nPlease enter a payroll number to search: ");
            searchQuery = keyboard.nextLong();

            searchForPayrollNumber(staffList, searchQuery);


        }while(!(searchQuery == 0));


    }

    private static void showDisplay(ArrayList<Personnel> staffList)
    {
        System.out.print("\n------------- CURRENT STAFF LIST -------------\n");
        for (Personnel person : staffList)
        {
            System.out.println("Payroll number: " + person.getPayNum());
            System.out.println("Surname: " + person.getSurname());
            System.out.println("First name(s): " + person.getFirstNames() + "\n");
        }
    }

    public static void searchForPayrollNumber(ArrayList<Personnel> staffList, long searchQuery)
    {
        long index = staffList.indexOf(searchQuery);;

        for (Personnel person: staffList)
        {
            if (person.getPayNum() == searchQuery)
            {               
                System.out.print("\n------------- Staff member found and removed! -------------");
                System.out.println("\n\nFirst Name(s): " + person.getFirstNames());
                System.out.println("\nSurname: " + person.getSurname());
                System.out.print("\n-----------------------------------------------");

                staffList.remove(index);
                return;
            }
        }

        System.out.print("\n------------- No staff members found. Program terminated -------------");
        return;

    }

}

package personnelPackage;

public class Personnel
{
    private long payrollNum;
    private String surname;
    private String firstNames;

    public Personnel(long payrollNum, String surname, String firstNames)
    {
        this.payrollNum = payrollNum;
        this.surname = surname;
        this.firstNames = firstNames;
    }

    public long getPayNum()
    {
        return payrollNum;
    }

    public String getSurname()
    {
        return surname;
    }

    public String getFirstNames()
    {
        return firstNames;
    }

    public void setSurname(String newName)
    {
        surname = newName;
    }
}

Answer 1

考虑使用迭代器进行搜索和删除：

Iterator<Personnel> i = staffList.iterator();
while (i.hasNext()) {
    Personnel p = i.next();
    if (p.getPayNum() == searchQuery) {
        // print message
        i.remove();
        return p;
    } 
}
return null;

如果严格要求使用list#remove()，则返回found人员p，并调用，如果（p！=null)stafflist.remove(p):

public static Personnel searchByPayNum(List<Personnel> ps, long num) {
    for (Personnel p : ps) {
        if (p.getPayNum() == num)
            return p;
    }
    return null;
}

在调用方代码中：

Personnel p = searchByPayNum(staffList, query);
if (p != null) {
    // log
    staffList.remove(p);
}