Question

package test;

   /* Read numbers from a text file and store them into an array; then
  sort the array and display it on the screen. Save the sorted array
in sorted.txt. */ 

import java.util.Scanner;
import java.io.*;

public class Test {

 static public void main(String[] args) throws IOException {

    /* Initialize the input stream reading from a text file */

Scanner inputFile; 
    inputFile = new Scanner(new File("arrayex1.txt"));

    /* Initialize the output stream writing into a text file */

PrintWriter outputFile;
outputFile = new PrintWriter(new FileWriter("sorted.txt"));

    /* Declare the array */

int[] numbers = new int[50];
int index = 0;

    int temp, smallest, smallest_index;

    /* Read the first number */
    numbers[index] = inputFile.nextInt();

    while (numbers[index] != 0) {

        index++;
        numbers[index] = inputFile.nextInt();

    }

    /* Sort the array using the selection sort method; the inner 
       loop finds the smallest unsorted number, and the outer
       loop places it in the right place. */ 

    for (int i = 0; i < index - 1; i++) {

        smallest = numbers[i];
        smallest_index = i;

        for (int j = i + 1; j < index; j++) {

        if (numbers[j] < smallest) { 
                smallest = numbers[j];
                smallest_index = j;
            }
        }

        /* If needed switch numbers[i] and numbers[smallest_index] */

        if (numbers[i] != numbers[smallest_index]) {

            temp = numbers[i];
            numbers[i] = numbers[smallest_index];
            numbers[smallest_index] = temp;
        }
    }

    /* Display the sorted array and also save the result in sorted.txt*/

    for (int i = 0; i < index; i++) {          
    System.out.print (numbers[i] + " ");
        outputFile.print (numbers[i] + " ");
    }

    inputFile.close();
    outputFile.close();

 }
  }

这是错误获取：

Exception in thread "main" java.util.NoSuchElementException
at java.util.Scanner.throwFor(Unknown Source)
at java.util.Scanner.next(Unknown Source)
at java.util.Scanner.nextInt(Unknown Source)
at java.util.Scanner.nextInt(Unknown Source)
at test.Test.main(Test.java:37)

请提供任何帮助和解决方案。提前谢谢。

Answer 1

始终通过调用Scanner.hasNext()检查扫描仪是否有下一个元素，而不是跳转到Scanner.Next()。

显示错误是因为扫描仪到达了最后一个元素，并且没有其他人再次显示。

Answer 2

首先，由于我猜测每次都不会有相同数量的数字排序，因此，我建议使用整数的ArrayList。

这是读入方法，隐含了导入：

    private static ArrayList<Integer> nums;
    public static void readIn() throws FileNotFoundException{
        Scanner sc = new Scanner(new File("file.txt"));
        boolean error = false;
        while(sc.hasNext()&&!error){
            try{
                nums.add(Integer.parseInt(sc.next()));
            }catch(NumberFormatException e){
                e.printStackTrace();
                error = true;
            }
        }
    }

这是最有效排序的代码。

        int[] numbers = (int[]) nums.toArray();
        Arrays.sort(numbers);
        System.out.println(Arrays.toString(numbers));

我的Java代码无法编译 - 我遇到了错误，我不明白为什么

2 个答案: