0 
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)

请任何帮助和解决方案将不胜感激。提前致谢。

4

2 回答 2

1

Scanner.hasNext()始终通过调用而不是跳入来检查 Scanner 是否有下一个元素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元素可以再次显示。

于 2012-12-09T18:41:36.360 回答
0

首先，由于我猜测每次排序的数字不会相同，因此，我建议使用整数的 ArrayList。

这是 read in 方法，并且隐含了导入：

    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;
            }
        }
    }

这可能是最有效排序的代码。

        int[] numbers = (int[]) nums.toArray();
        Arrays.sort(numbers);
        System.out.println(Arrays.toString(numbers));
于 2014-06-23T17:46:15.717 回答