Question

我需要帮助解决以下问题。首先我列出了Numbers.java，下面是我对Strings.java的尝试。请帮忙！

编写一个类似于Numbers.java的程序Strings.java，它读入一个String对象数组并对它们进行排序。您可以只复制和编辑Numbers.java。

package hw05;

/*  
Demonstrates selectionSort on an array of integers.
*/ 

import java.util.Scanner;


public class Numbers {  
// --------------------------------------------  
// Reads in an array of integers, sorts them,  
// then prints them in sorted order.  
// --------------------------------------------  
public static void main(String[] args) {

    Integer[] intList;
    int size;

    Scanner scan = new Scanner(System.in);

    System.out.print("\nHow many integers do you want to sort? ");
    size = scan.nextInt();
    intList = new Integer[size];

    System.out.println("\nEnter the numbers...");
    for (int i = 0; i < size; i++)
        intList[i] = scan.nextInt();
    Sorting.selectionSort(intList);

    System.out.println("\nYour numbers in sorted order...");
    for (int i = 0; i < size; i++)
        System.out.print(intList[i] + " ");
    System.out.println();

}
}

My Strings课程

package hw05;

/*
Demonstrates selectionSort on an array of strings.
*/
import java.util.Scanner;

public class Strings {
    // --------------------------------------------
    // Reads in an array of strings, sorts them,
    // then prints them in sorted order.
    // --------------------------------------------
    public static void main(String[] args) {

        String[] stringList;
        String size;

        Scanner scan = new Scanner(System.in);

        System.out.print("\nHow many strings do you want to sort? ");
        size = scan.nextLine();
        stringList = new String[size];

        System.out.println("\nEnter the strings...");
        for (String i = 0; i < size; i++)
            stringList[i] = scan.nextLine();
        Sorting.selectionSort(stringList);

        System.out.println("\nYour strings in sorted order...");
        for (String i = 0; i < size; i++)
            System.out.print(stringList[i] + " ");
        System.out.println();

    }
}

Answer 1

由于此声明，您将收到错误：

String size;

正如您稍后将其用作整数：

stringList = new String[size];

你应该像这样使用Integer.parse：

stringList = new String[Integer.parseInt(size)];

但是你之后再次使用大小的循环，所以我建议将它恢复为你在Numbers.java中的int。务必将扫描仪更改回：

scan.nextInt();

除此之外，我看不到任何其他问题。也许如果你发布输出并指出什么不起作用？

Java中的多态排序转换

1 个答案: