Question

这是我第一次为字符串写一个冒泡排序，显然我遇到了很多错误而程序无法运行。我不知道如何解决它。我的代码是：

import java.util.*;
    public class SortingRecord{
        public static void main(String args[]){
            Scanner kb = new Scanner(System.in);
            System.out.println("How many people?");
            int n = Integer.parseInt(kb.nextLine());
            Record[] records = new Record[n];
            for(int i = 0; i<n; i++){
                System.out.println("Inputting record["+i+"]:");
                System.out.print("Please input <First Name>:");
                String firstName = kb.nextLine();
                System.out.println("Please input <Last Name>:");
                String lastName = kb.nextLine();
                records[i] = new Record(firstName, lastName);
            }
            sort(records);
            System.out.println("----------------");
            System.out.println("Print name in dictinary order:");
            for(int i = 0; i < n ; i++)
                System.out.println();
         } 

         public static void sort(Record[] records){
             if (records == null || records.length <= 1) return;
             int n = records.length;

             for(int i = 0; i< records.length ; i++){
                 for(int j = i+1 ; j< records.length; j++){

无法找到符号方法compareTo（Record）。

                     if(records[j] .compareTo(records[i]) < 0){

它说Record无法转换为java.lang.String

                     String temp = records[i];
                     records[i] = records[j];
                     records[j] = temp;
                 }
              }
              System.out.println(records[i]);

             }
         }

    }

  class Record{
      public String firstName = "";
      public String lastName = "";
      public Record(String firstName, String lastName){
         this.firstName = firstName;
         this.lastName = lastName;
      }
 }

Answer 1

让我们来看看这个明显的错误：

if (records[j].compareTo(records[i]) < 0) {

Record不提供任何compareTo方法，因此您无法调用它 - 它不存在。

下一个错误：

String temp = records[i];

因为Record不是String的类型，因此无法分配，显而易见的解决方案是使用Record，而不是...... < / p>

Record temp = records[i];
records[i] = records[j];
records[j] = temp;

好的，但我们如何解决compareTo问题？这比听起来更复杂，当您实施Comparable interface（或直接实施compareTo方法）时，我不会选择此路径。为什么？因为您可能想要更改对记录进行排序的方式，并且实现该方法会将您锁定为单个用例。

相反，我使用传入方法的Comparator进行实际比较，为调用者提供了更改比较实际工作方式的灵活性

public static void sort(Record[] records, Comparator<Record> comparator) {
    if (records == null || records.length <= 1) {
        return;
    }
    int n = records.length;

    for (int i = 0; i < records.length; i++) {
        for (int j = i + 1; j < records.length; j++) {
            if (comparator.compare(records[j], records[i]) < 0) {
                Record temp = records[i];
                records[i] = records[j];
                records[j] = temp;
            }
        }
        System.out.println(records[i]);

    }
}

然后你可以做一些像......

sort(records, new Comparator<Record>() {
    @Override
    public int compare(Record o1, Record o2) {
        return o1.firstName.compareTo(o2.firstName);
    }
});

或

sort(records, new Comparator<Record>() {
    @Override
    public int compare(Record o1, Record o2) {
        return o1.lastName.compareTo(o2.lastName);
    }
});

甚至

sort(records, new Comparator<Record>() {
    @Override
    public int compare(Record o1, Record o2) {
        int compare = o1.firstName.compareTo(o2.firstName);
        if (compare == 0) {
            compare = o1.lastName.compareTo(o2.lastName);
        }
        return compare;
    }
});

或者您可能需要的其他组合才能满足您的要求

我建议您查看Comparator了解更多详情

我还应该指出，您也可以使用Collections对象，但是您需要将其转换为List而不是数组...

Collections.sort(Arrays.asList(records), new Comparator<Record>() {...});

程序无法按字典顺序输出名称;（

对我来说很好......

import java.util.Comparator;

public class Test {

    public static void main(String[] args) {
        new Test();
    }

    public Test() {
        Record[] records = new Record[] {
            new Record("B", "B"),
            new Record("C", "B"),
            new Record("D", "B"),
            new Record("A", "E"),
            new Record("A", "B"),
            new Record("A", "C"),
            new Record("A", "A"),
        };
        sort(records, new Comparator<Record>() {
                 @Override
                 public int compare(Record o1, Record o2) {
                     int compare = o1.firstName.compareTo(o2.firstName);
                     if (compare == 0) {
                         compare = o1.lastName.compareTo(o2.lastName);
                     }
                     return compare;
                 }
             });

        for (Record record : records) {
            System.out.println(record);
        }
    }

    public static void sort(Record[] records, Comparator<Record> comparator) {
        if (records == null || records.length <= 1) {
            return;
        }

        for (int i = 0; i < records.length; i++) {
            for (int j = i + 1; j < records.length; j++) {
                if (comparator.compare(records[j], records[i]) < 0) {
                    Record temp = records[i];
                    records[i] = records[j];
                    records[j] = temp;
                }
            }
        }
    }

    class Record {

        public String firstName = "";
        public String lastName = "";

        public Record(String firstName, String lastName) {
            this.firstName = firstName;
            this.lastName = lastName;
        }

        @Override
        public String toString() {
            return firstName + " " + lastName;
        }


    }
}

输出

A A
A B
A C
A E
B B
C B
D B

Answer 2

你在Record类中没有 compareTo 方法，这就是为什么找不到它:)你可能应该实现Comparable接口。< / p>

至于＆＃34;记录无法转换为java.lang.String＆＃34;，使用 toString 方法，你可以转换它，虽然你可能想要覆盖的toString

还请看一下这个例子： Why should a Java class implement comparable?

字符串的冒泡排序。 java的

2 个答案: