从HashSet数组中按索引获取项目，并在列表中显示每个项目的平均长度 [英] Get an item by index from the HashSet array and also display the average length of each item in the list

问题描述

我创建了使用HashSet数组的代码。我是Java编程的新手，我想知道如何完成这两个任务：

Hi I have created code that uses a HashSet array. I am brand new to Java programming and I would like to know how to accomplish these two tasks:

*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

*Get an item by index from the HashSet array *display the average length of each item in the list

我的代码根本不长，所以我将整个代码粘贴到这里。

My code isn't long at all so I am pasting the entire code here. Thanks for all your help.

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();

            }


        }

    });


    }

}

推荐答案

集合不是您想要的，因为：

A Set is not what you want, because:

• 集合仅容纳唯一值


• 集合中的元素没有顺序，因此您无法按索引访问元素

一个更好的选择是列表，例如 ArrayList ，它允许任何（即重复）值，并且可以通过使用<$ c的索引进行访问$ c> list.get（i）。

A better choice is a List, such as an ArrayList, which allows any (ie duplicate) values and can be accessed by index using list.get(i).

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