Question

我有点迷失在我错误的地方这里是我的主要文件：

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，因为当它们按字母顺序排列时，第一个对象成为第一个名称，最后一个对象成为姓氏。

这是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，最终输出是用户按照输入的顺序输入的所有名称。同样，名字甚至不按字母顺序排列。任何帮助都会非常感激！

Answer 1

这是一个简化且正确运行的代码版本（如果我很好地理解了该主要方法的目标）：

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]);
}

很少考虑：

您不需要TreeSort对数组进行排序，只需使用Arrays.sort方法即可。还要注意将其作为输入限制为开始和结束的其他参数，这些参数不一定与数组的完整初始大小匹配（因为您从控制台输入填充它）
要填充数组的不同元素，需要移动到下一个索引，因此count变量需要在每次新迭代时递增1
一旦对数组进行排序，您不需要第一个和最后一个元素的临时变量，只需使用第一个索引（0）和最后一个填充（count - 1）
您还需要将整个循环限制为数组的最大大小，以免掉出其大小并获得异常

希望您更好地了解循环，数组，排序以及如何改进代码。

更新：上述代码将在每次输入后逐个读取名称，直到提供空输入或达到MAX_SIZE限制。如果你想一次性输入所有名字，你甚至不需要循环，你的代码将会进一步简化，但你应该以某种方式分离标记（你的名字），否则它们都将被视为一个大名。查看下面的代码，它使用String.split（）方法分隔标记（名称），并使用“”（空格）作为标记分隔符。

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

您可以使用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以排序顺序存储值，first和last方法分别返回set中的第一个和最后一个值。

现在来看你的代码：

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个具有相同初始值的对象，即当它为第二遍传递时，所有对象都将包含该第二个值，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。 / p>

    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同一个数组元素（相同的索引）十次。在此之后，您将此对象分配给第一个和最后一个变量。因此，在所有此迭代之后，第一个和最后一个将包含用户输入的最后一个元素。而且compareTo方法返回无处捕获的整数，并且当你使用treeset时比较是没用的（在这种情况下）。

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，你要么做其他事情，要么你使用了错误的逻辑。在树集中插入名称，并使用first和last方法获取名称。

Java：当用户提供的名称按字母顺序排列时，仅打印First和Last

2 个答案: