Question

我将这段代码放在一起对数字进行排序。它可以工作，如果可以的话，我想变得更短，更干净。我正在使用两个扫描仪来扫描同一文件。我不太确定如何使用方法来接受输入文件。

import java.io.*;


public class test {
    public static void main(String[] args) throws Exception {
        Scanner scan = new Scanner(new File("/Users/WalterCueva/Desktop/4.txt")); //provide file name from outside
        int counter = 0; //keep track of how many iargs[0]ntegers in the file

        while (scan.hasNextInt()) {
            counter++;
            scan.nextInt();
        }
        Scanner scan2 = new Scanner(new File("/Users/WalterCueva/Desktop/4.txt"));
        int[] numbers = new int[counter];
        for (int i = 0; i < numbers.length; i++) {
           numbers[i] = scan2.nextInt(); //fill the array with the integers
        }
        for (int i = 0; i < numbers.length; i++) {
            System.out.println(numbers[i]);
            ; //fill the array with the integers
        }

        new test().sort(numbers);
        System.out.println(Arrays.toString(numbers));
    }

    public void sort(int[] data) {

        for (int i = 0; i < data.length - 1; i++) {
            for (int j = 0; j < data.length - 1 - i; j++) {
                // do the swap if required
                if (data[j] > data[j+1]) {
                    int tmp = data[j+1];
                    data[j+1] = data[j];
                    data[j] = tmp;
                }
            }
        }
    }
}

Answer 1

替换这部分代码：

int[] numbers = new int[counter];
for (int i = 0; i < numbers.length; i++) {
    numbers[i] = scan2.nextInt(); //fill the array with the integers
}
for (int i = 0; i < numbers.length; i++) {
    System.out.println(numbers[i]);
    ; //fill the array with the integers
}
new test().sort(numbers);

与此：

int[] numbers = new int[counter];
for (int i = 0; i < numbers.length; i++) {
    numbers[i] = scan2.nextInt(); //fill the array with the integers
    System.out.println(numbers[i]);
}
Arrays.sort(numbers);

并放下public void sort(int[] data)函数。

Answer 2

我相信这是最短的时间。

首先，我们使用List，它会随着添加更多数据而动态增长。这样可以防止您需要对数据进行两次迭代。

第二，我们使用内置函数Collections.sort()，该函数使用Tim-sort。 Tim sort能够以最坏的时间复杂度N log (N)对数据进行排序（与您的bubbleort实现N²相比）。

import java.io.File;
import java.util.*;

public class test {
    public static void main(String[] args) throws Exception {
        try (Scanner scan = new Scanner(new File("text.txt"))) {
            List<Integer> list = new ArrayList<Integer>();
            while (scan.hasNextInt())
                list.add(scan.nextInt());
            Collections.sort(list);

            System.out.println(list);
        }
    }
}

Answer 3

通过使用纯Java 8内容，可以更轻松地完成此操作。看看这个：

public class TestClass {

    private static final String FILE_LOCATION = "C:\\tmp\\file.txt";

    public static void main(String[] args) {
        List<Integer> lines = new ArrayList<>();
        try (Stream<String> stream = Files.lines(Paths.get(FILE_LOCATION))) {
            lines = stream.map(Integer::parseInt).collect(Collectors.toList());
        } catch (IOException e) {
            e.printStackTrace();
        }
        lines.sort(Comparator.naturalOrder());
        lines.forEach(System.out::println);
    }

}

Answer 4

 import java.io.*;
    import java.util.*;
    public class test {
     public static void main(String[] args) throws Exception {

         File file = new File ("/Users/WalterCueva/Desktop/4.txt");
          if (!file.exists()) {
                System.out.println("The file does not exist ");
                System.exit(0);
            }

         Scanner input = new Scanner(file);

        ArrayList <Integer>numbers = new ArrayList<>();


  while(input.hasNext()) {
            numbers.add( input.nextInt()); //fill the array with the integers
        }

       input.close();

       for (double e: numbers) {
         System.out.println(e);
       }

    java.util.Collections.sort(numbers);
    System.out.println(numbers);

    }}

如何使这段代码更短，更清晰

4 个答案: