Java：当用户给出的名字按字母顺序排列时，只打印名字和姓氏

Question

我有点迷失在我哪里出错了这里是我的主文件：

import java.util.Scanner;
import java.util.Arrays;
import java.util.*; 
import java.util.TreeSet; 

public class personSorter 
{
  public static void main(String[] args)
  {
  int count = 0; 
  Scanner in = new Scanner(System.in);

  boolean more = true;
  Person first = null;
  Person last = null; 

  Person[] people= new Person[10]; //my array 

  while(more)
  {
  System.out.println("Please enter the person's name or a blank line to quit");
  String names = in.nextLine(); 

  if (names.equals(""))
     {    
     more = false;
     }
 else
 {
        Person p1 = new Person(names);  //creating 10 person objects to be used
        Person p2 = new Person(names);
        Person p3 = new Person(names);
        Person p4 = new Person(names);
        Person p5 = new Person(names);
        Person p6 = new Person(names);
        Person p7 = new Person(names);
        Person p8 = new Person(names);
        Person p9 = new Person(names);
        Person p10 = new Person(names); 

       people[count] = p1; // using my person objects and declaring the index of variable count
       people[count] = p2;
       people[count] = p3;
       people[count] = p4;
       people[count] = p5;
       people[count] = p6;
       people[count] = p7;
       people[count] = p8;
       people[count] = p9;
       people[count] = p10;

        first = people[count];
        last  = people[count];

    TreeSet<String> treeSet = new TreeSet<String>(); //using TreeSort to get the names entered by user in ascending order

     treeSet.add(names);
     treeSet.add(names);
     treeSet.add(names);
     treeSet.add(names);
     treeSet.add(names);
     treeSet.add(names);
     treeSet.add(names);
     treeSet.add(names);
     treeSet.add(names);
     treeSet.add(names);



   first.compareTo(p1); //after TreeSort first compare method called of first object as it is now the first name in alphabet order
   last.compareTo(p10); //after TreeSort last compare method called of last object as it is now the last name in alphabet order

  count++;       
  }
}

//printing out the first and last name(in alphabet order) of the entered names 
  System.out.println("First: " + first.toString()); 
  System.out.println("Last: " + last.toString());
 } 
 }

我使用 TreeSet 按字母顺序排列名称。然后我在对象 1 和 10 上使用了对 compareTo 方法的调用，只是因为当它们按字母顺序排列时，第一个对象成为名字，最后一个对象成为姓氏。

这里是 Person.java:

public class Person implements Comparable <Person> 
    {
        private String name;

public Person(String n) 
{
    name = n;
}

public String getName() 
{
    return name;
}

   @Override
public int compareTo(Person others) 
{
    if (name.compareTo(others.name) == 1) 
    {
        return 0;
    } 
    else if (name.compareTo(others.name) < 0) 
    {
        return -1;
    } 
    else 
    {
        return 1;
    }
   }
   public String toString() 
   {
   return "[" + name + "]";
   }

}

但是，最终输出的是用户输入的所有名称，按照输入的顺序排列。就像，名字甚至不按字母顺序排列。任何帮助将不胜感激！

Answer 1

这是您的代码的简化且正确工作的版本（如果我很好地理解该主要方法的目标）：

public static void main(String[] args) {
    int count = 0;
    Scanner in = new Scanner(System.in);

    boolean more = true;
    int MAX_SIZE = 10;

    Person[] people = new Person[MAX_SIZE]; // my array

    while (more && count < MAX_SIZE) {
        System.out.println("Please enter the person's name or a blank line to quit");
        String name = in.nextLine();

        if (name.equals("")) {
            more = false;
        } else {
            Person p = new Person(name); 
            people[count++] = p;
        }
    }
    in.close();

    Arrays.sort(people, 0, count);

    // printing out the first and last name(in alphabet order) of the entered names
    System.out.println("First: " + people[0]);
    System.out.println("Last: " + people[count - 1]);
}

几个注意事项：

您不需要 TreeSort 来对数组进行排序，只需使用 Arrays.sort 方法即可。还要注意它作为输入将其限制为开始和结束的更多参数，这些参数不一定与数组的完整初始大小匹配（因为您是从控制台输入填充它的）
要填充数组的不同元素，您需要移动到它的下一个索引，因此计数变量需要在每次新迭代时递增 1
数组排序后，第一个和最后一个元素不需要临时变量，只需使用第一个索引 (0) 和最后一个填充的索引 (count - 1)
您还需要将整个循环限制为数组的最大大小，以免超出其大小并引发异常

希望您能更好地理解循环、数组、排序以及如何改进您的代码。

Update：上面的代码会在每次输入后一个一个地读取名称，直到提供一个空输入或达到MAX_SIZE限制。如果你想一次输入所有名字，你甚至不需要循环，你的代码将进一步简化，但是你应该以某种方式将标记（你的名字）分开，否则它们将被视为一个大名字. 查看下面的代码，它使用 String.split() 方法并使用“”（space）作为标记分隔符来分隔标记（名称）。

public static void main(String[] args) {
    int count = 0;
    Scanner in = new Scanner(System.in);
    int MAX_SIZE = 10;
    Person[] people = new Person[MAX_SIZE];

    System.out.println("Please enter the person's name or a blank line to quit");
    String line = in.nextLine();

    String[] names = line.split(" ");
    for (int i = 0; i < MAX_SIZE; i++) {
        Person p = new Person(names[i]);
        people[count++] = p;
    }
    in.close();

    Arrays.sort(people, 0, count);

    // printing out the first and last name(in alphabet order) of the entered names
    System.out.println("First: " + people[0]);
    System.out.println("Last: " + people[count - 1]);
}

Answer 2

您可以对 TreeSet.Read java 文档中的 first() 和 last() 方法执行相同的操作以获取详细信息。 http://docs.oracle.com/javase/7/docs/api/java/util/TreeSet.html

TreeSet 按排序顺序存储值，first 和 last 方法 return 分别存储集合中的第一个和最后一个值。

现在进入您的代码：

else
{
    Person p1 = new Person(names);  //creating 10 person objects to be used
    Person p2 = new Person(names);
    Person p3 = new Person(names);
    Person p4 = new Person(names);
    Person p5 = new Person(names);
    Person p6 = new Person(names);
    Person p7 = new Person(names);
    Person p8 = new Person(names);
    Person p9 = new Person(names);
    Person p10 = new Person(names);

您正在创建 10 个具有相同初始化值的对象，即当它循环第二次时，所有对象都将包含第二个值，并且初始值将丢失。在这种情况下，您的对象将仅包含最后输入的值.

    people[count] = p1; // using my person objects and declaring the index of variable count
    people[count] = p2;
    people[count] = p3;
    people[count] = p4;
    people[count] = p5;
    people[count] = p6;
    people[count] = p7;
    people[count] = p8;
    people[count] = p9;
    people[count] = p10;

    first = people[count];
    last  = people[count];

然后您将这些相同的对象分配给相同的数组元素（相同的索引）十次。在此之后，您将此对象分配给 first 和 last variable.So，因为所有此迭代 first 和 last 将包含用户输入的最后一个元素。并且 compareTo 方法 returns 整数在您使用 treeset 时（在本例中）是无处捕获和比较的。

first.compareTo(p1); //after TreeSort first compare method called of first object as it is now the first name in alphabet order
last.compareTo(p10); //after TreeSort last compare method called of last object as it is now the last name in alphabet order

总而言之，要么您正在尝试做其他事情，要么您使用了错误的逻辑。在树集中插入名称并使用 first 和 last 方法获取名称。

Java：当用户给出的名字按字母顺序排列时，只打印名字和姓氏

Java: Printing only the First and Last when the names that the user gave are organized alphabetically

java

treeset