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好的,所以我正在研究一种对二维数组进行排序的方法,其中一个维度有一个字符串,另一个维度有一个 int(为方便起见,存储为字符串)我到处寻找关于如何对数组进行排序的解决方案一种将 firstArray[1] 中的数据同时移动的方式(其索引是移动的子项:)作为 firstArray[0]

这个效果是通过使用这个来实现的

Arrays.sort(fps, new Comparator<String[]>() {
            @Override
            public int compare(final String[] entry1, final String[] entry2) {
                final String time1 = entry1[0];
                final String time2 = entry2[0];
                return time1.compareTo(time2);
            }
        });

现在我在这件事上遇到了麻烦。我将在这里逐步完成代码,如果您发现它有问题,请告诉我。

首先我有数组:

String[][] fps = new String[2][15];
Arrays.fill(fps[0], "empty");
Arrays.fill(fps[1], "0");

其次,我用程序的另一部分给我的一些东西填充数组,对于这个例子,我将使用垃圾值:

fps[0][0] = "Java";
fps[1][0] = "1";

fps[0][1] = "C++";
fps[1][1] = "14";

fps[0][2] = "C#";
fps[1][2] = "21";

fps[0][3] = "Python";
fps[1][3] = "9001";

现在是我调用上述排序命令的地方(请注意,这些值不会完全填充数组,有些 bin 没有新数据。)

Arrays.sort(fps, new Comparator<String[]>() {
            @Override
            public int compare(final String[] entry1, final String[] entry2) {
                final String time1 = entry1[0];
                final String time2 = entry2[0];
                return time1.compareTo(time2);
            }
        });

现在我们对数组进行了排序,我想在二维数组中搜索一个值,所以我使用 Arrays.search 来查找查询所在的 bin。

int searchIndex = Arrays.binarySearch(fps[0], "Java");
System.out.println(searchIndex);

这就是代码,我认为我已经将问题隔离为排序部分无法正常工作。如果你们还有任何问题,请在评论中发表。同样,如果您对这个令人费解的问题有可能的解决方案,我很想听听!


PS:为了清楚我有这个工作,然后我关闭了我的lappy,下次我启动(从那时起)它没有工作。

PPS:根据要求输出:

电流输出:

-16
FPS:
    0 ---- No FPS For you!
    1 ---- Only one FPS
    2 ---- Only two FPS
    3 ---- Only three FPS
    4 ---- Only four FPS
    5 ---- Only five FPS
    6 ---- Only six FPS
    7 ---- Only seven FPS
    8 ---- Only eight FPS
    9 ---- Only nine FPS
    1 ---- Blah!

预期/希望的输出:

-16
FPS:
    1 ---- Blah!
    0 ---- No FPS For you!
    8 ---- Only eight FPS
    5 ---- Only five FPS
    4 ---- Only four FPS
    9 ---- Only nine FPS
    1 ---- Only one FPS
    7 ---- Only seven FPS
    6 ---- Only six FPS
    3 ---- Only three FPS
    2 ---- Only two FPS

PPPS:如果您想查看我目前正在使用的代码:

import java.util.*;

public class Test
{
  public static void main (String [] args)
  {

    String[][] fps = new String[2][15];
    Arrays.fill(fps[0], "empty");//Fill up all the spaces so the sort and the search dont crap out
    Arrays.fill(fps[1], "0"); 


    //fps[ROW][COLOUMN] = Value + "";
    //fps[ROW][COLOUMN] = Value Instances + "";

    fps[0][0] = "No FPS For you!";
    fps[1][0] = 0 + "";

    fps[0][1] = "Only one FPS";
    fps[1][1] = 1 + "";

    fps[0][2] = "Only two FPS";
    fps[1][2] = 2 + "";

    fps[0][3] = "Only three FPS";
    fps[1][3] = 3 + "";

    fps[0][4] = "Only four FPS";
    fps[1][4] = 4 + "";

    fps[0][5] = "Only five FPS";
    fps[1][5] = 5 + "";

    fps[0][6] = "Only six FPS";
    fps[1][6] = 6 + "";

    fps[0][7] = "Only seven FPS";
    fps[1][7] = 7 + "";

    fps[0][8] = "Only eight FPS";
    fps[1][8] = 8 + "";

    fps[0][9] = "Only nine FPS";
    fps[1][9] = 9 + "";
    /* FUMBLE WITH ARRAY AFTER THIS LINE ONLY!*/

    //Things to have inputed into the method:
    //currentValue (from the line)
    //currentVariable (so we know what the name of the array we're dealing with is named)

    String currentValue = "Blah!"; //This is the value that will be searched for in the array, if found its child int is incremented by one, if not found it is added to the array.


    //Do a binary sort then search in the fps[0] side of things, makesure that the [1] are carried with the [0] changes.

    Arrays.sort(fps, new Comparator<String[]>() {
            @Override
            public int compare(final String[] entry1, final String[] entry2) {
                final String time1 = entry1[0];
                final String time2 = entry2[0];
                return time1.compareTo(time2);
            }
        });

    int searchIndex = Arrays.binarySearch(fps[0], currentValue); //Get the index of the current search value
    System.out.println(searchIndex); //  <-- Returns a neg number each time which shows that the sorting is not working correctly, and therefore the search is being thrown off... need to somehow fix.

    if(searchIndex >= 0)// If the value is found
    {
      fps[0][searchIndex] = (Integer.parseInt(fps[0][searchIndex]) + 1) + "";  //Add one instance to the value

    } //end if
    else //Otherwise find the next open value spot and change it to the current search query (and assign its instances to 1
    {

      for(int i = 0; i < fps[1].length ; i++)
      {
        if(fps[1][i].equals("empty"))
        {
          fps[1][i] = currentValue;
          fps[0][i] = 1 + "";
          i = fps[1].length; //force the for loop to exit

          Arrays.sort(fps, new Comparator<String[]>() {
            @Override
            public int compare(final String[] entry1, final String[] entry2) {
                final String time1 = entry1[0];
                final String time2 = entry2[0];
                return time1.compareTo(time2);
            }
        }); //end Arrays.sort
        }//end if
      }//end for
    }//end else


    //... Print array in rectangular form
    System.out.println("FPS:");

    for (int i =0; (!(fps[1][i].equals("empty")) ) ; i++)
    {
      System.out.println("\t" + fps[0][i] + " ---- " + fps[1][i] );


    }//end for
  }//end main
}//end class
4

3 回答 3

1

我相信你的索引是倒退的。你正在排序fpsfps只有 2 个正在排序的元素。我相信您正在尝试对 15 个元素进行排序。如果您反转索引,我相信您将获得所需的排序。

String[][] fps = new String[15][2];

在这种情况下,您可能会考虑对象数组而不是二维数组。它似乎是一个更合乎逻辑的结构,可以避免这种混淆。

于 2013-03-08T21:37:09.803 回答
0

除了这个Answer指出的问题外,这里还有一个问题:

int searchIndex = Arrays.binarySearch(fps[0], "Java");

由于您使用自定义比较器进行排序,因此您需要使用相同的自定义比较器进行二分搜索。使用binarySearch(T[] a, T key, Comparator<? super T> c). (如果你使用 2-arg 版本,你应该得到一个异常,因为String[]它没有实现Comparable。)

于 2013-03-09T02:47:26.573 回答
0

你有关于这些对象的任何资源吗(在与此类似的实例中使用时

有关在使用自定义对象时如何执行此操作的示例,请参阅:使用 Comparator 降序排序(用户定义的类) 。

一个维度有一个字符串,然后另一个维度有一个 int (为了方便起见,存储为字符串)

它不方便,因为对数字的字符串表示形式进行排序与​​对数字进行排序不同。使用自定义对象将允许您正确存储数据,以便您可以进行适当的排序。

于 2013-03-09T02:33:07.740 回答