关于Java：如何使数组列表的get方法更简单？

How do I make this get method for array lists simpler?

本问题已经有最佳答案，请猛点这里访问。

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import java.util.*;

public class ChristmasParty
{
private Scanner input = new Scanner(System.in);

public static void main(String[] args)
{
ChristmasParty cp = new ChristmasParty();
cp.run();
}

public void run()
{
menu();
addName();
}

public void menu()
{
System.out.println("Welcome to the guests lists program");
System.out.println("Enter a name or enter X to quit");
}

public void addName()
{
String nameAdded ="";
ArrayList<String> guestsLists = new ArrayList<String>();
do
{
System.out.print("Enter a name:");
nameAdded = input.nextLine();

guestsLists.add(nameAdded);
System.out.println("-----------------------------------------------");
System.out.println(nameAdded +" has been added to the guests list.");

}while(!nameAdded.equals("X"));
System.out.println("Guests lists:");
System.out.println("========================");
System.out.println(guestsLists.get(0));
System.out.println(guestsLists.get(1));
System.out.println(guestsLists.get(2));
System.out.println(guestsLists.get(3));
System.out.println(guestsLists.get(4));
System.out.println(guestsLists.get(5));
System.out.println(guestsLists.get(6));
System.out.println(guestsLists.get(7));
System.out.println(guestsLists.get(8));
System.out.println(guestsLists.get(9));
System.out.println(guestsLists.get(10));
System.out.println(guestsLists.get(11));
System.out.println(guestsLists.get(12));
System.out.println(guestsLists.get(13));
System.out.println(guestsLists.get(14));
System.out.println(guestsLists.get(15));
}
}

您好，我正在尝试编写一个代码，它将提示用户输入一个名称，该名称将保存在ArrayList中，然后一旦用户退出并按"x"退出，它将显示列表中的所有名称。但是，如何使system.out.println(guessLists.get())；代码比全部输入简单？

放这个代码：

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}while(!nameAdded.equals("X"));

System.out.println("Guests lists:");
System.out.println("========================");

for(int i = 0; i < guestsLists.size(); i++)
{
System.out.println(guestsLists.get(i));
}

或

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for(String s : guestsLists)
{
System.out.println(s);
}

然后移除

system.out.println("================")；