Question

我正在努力做一些功课。除了这个问题，我已解决了所有问题。在搜索网站后，我决定只问自己的问题。

该代码旨在制作一个小菜单，询问用户是否要将电子邮件输入数据库，搜索以前输入的电子邮件或退出。

当我尝试进行二进制搜索以确定电子邮件是否已存在于数据库中时，会出现问题。

错误发生在addNewEmail方法的第二行（最后一行）。提前感谢您的帮助。

import java.util.Scanner;
import java.util.ArrayList;
import java.util.Collections;

public class project14_Brady {

    private static final int ADD = 1;
    private static final int SEARCH = 2;
    private static final int QUIT = 3;

    public static void main(String[] args) 
    {

        ArrayList<String> emailAddresses = new ArrayList<String>();
        int size = 0;

        Scanner input = new Scanner(System.in);

        // Priming read
        int menu = menuChoice(input);
        while(menu != QUIT)
        {
            if (menu == ADD)
            {
                System.out.println("Enter the email address");
                String email = input.nextLine();
                size = addNewEmail(emailAddresses, size, email);
            }
            else if (menu == SEARCH)
            {
                String email = autoComplete(emailAddresses, size, input);
                if (email != null)
                    System.out.println("Found: " + email);
                else
                    System.out.println("No matching email was found");
            }
            else
            {
                System.out.println("Unanticipated case");
            }

            // priming read
            menu = menuChoice(input);

        } // end while

    }

    public static String autoComplete(ArrayList<String> data, int size, Scanner input)
    {
        System.out.println("Enter the first letters, one at a time");
        String start = "";

        while (true)
        {
            String read = input.nextLine();
            start += read;

            System.out.println("DEBUG: " + start);

            int count=0;
            String result="";

            for (int i=0; i<size; ++i)
            {
                System.out.println("DEBUG: " + data.get(i));
                if (data.get(i).startsWith(start))
                {
                    result = data.get(i); // keep just the last one
                    System.out.println(data.get(i));
                    ++count;
                }

            }

            if (count == 1)
                return result;

            if (count == 0)
                return null;

        }
    }

    public static int menuChoice(Scanner keyboard)
    {
        System.out.println("Please choose from the following menu of choices:");
        System.out.println("1. Enter a new email address");
        System.out.println("2. Find an existing email address");
        System.out.println("3. Quit.");
        System.out.println("What is your choice?");

        int choice = keyboard.nextInt();
        keyboard.nextLine(); // get rid of newline

        // Allow the user to re-enter data
        while (choice < ADD || choice > QUIT)
        {
            System.out.println("You must choose a value between 1 and 3");
            System.out.println("Please re-enter your choice");
            choice = keyboard.nextInt();
            keyboard.nextLine();
        }

        return choice;
    }

    public static int addNewEmail(ArrayList<String> data, int size, String insertMe)
    {
        if (data.binarySearch(data, 0, size, insertMe) > 0)
        {
            System.out.println("That email address has already been inserted");
            return size; // already in array
        }

        if (size == data.size())
        {
            System.out.println("Too many addresses are stored");
            return size; // the array is full already
        }

        // This is essentially one inner loop of insertion sort
        int index;
        for (index = size; index > 0 && data.get(index-1).compareTo(insertMe) > 0; --index)
        {
            data.add(index, data.get(index-1));
        }

        data.add(index, insertMe);

        return size+1;
    }

}

Answer 1

ArrayList没有binarySearch方法。你可以试试这个

if (Collections.binarySearch(data, insertMe) > 0)

这是编译，但我没有详细阅读你的代码，说它绝对是你想要做的。

对于ArrayList <string>类型，方法binarySearch（ArrayList <string>，int，int，String）是未定义的？

1 个答案: