从HashSet数组中按索引获取项目,并显示列表中每个项目的平均长度

Question

嗨,我已经创建了使用HashSet数组的代码.我是Java编程的新手,我想知道如何完成这两项任务:

*从HashSet数组中按索引获取项目*显示列表中每个项目的平均长度

我的代码不长,所以我在这里粘贴整个代码.感谢你的帮助.

public class MainActivity extends Activity {

Button aButton; // Global Scope
Button sButton;
TextView text2;
EditText eText;
HashSet<String> list = new HashSet<String>();


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.new_layout); 

    aButton = (Button) this.findViewById(R.id.button1);
    text2 = (TextView) this.findViewById(R.id.textView1);

    //Clickable Saved Input will display alert
    text2.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View v) {
            Toast.makeText(getApplicationContext(), "Complete!", Toast.LENGTH_LONG).show();

        }
    });

    aButton.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View v) {


            list.add("Books");
            list.add("Newspapers");
            list.add("Magazines");
            String listString = "";

            for (String s : list) {
                listString += s + " - ";
            }
            text2.setText(listString);
            }
        });




    sButton = (Button) this.findViewById(R.id.button2);
    eText = (EditText) this.findViewById(R.id.editText1);

    sButton.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View view) {


             //Log.v("EditText", eText.getText().toString());
            if( !list.add(eText.getText().toString()) )
            {
                System.out.println("Not Unique Item");
                Toast.makeText(getApplicationContext(), "Already Saved!", Toast.LENGTH_LONG).show();
            } else 
            {
                System.out.println("Unique Entry Added");
                Toast.makeText(getApplicationContext(), "Saved To Items.", Toast.LENGTH_LONG).show();

            }


        }

    });


    }

}

Answer 1

套装不是你想要的,因为:

• 设置仅保存唯一值
• Set的元素没有顺序,因此您无法通过索引访问元素

更好的选择是List,例如a ArrayList,它允许任何(即重复的)值,并且可以通过索引使用list.get(i).