同时排序两个数组

Question

I'm learning and understanding Java now, and while practising with arrays I had a doubt. I wrote the following code as an example:

class example
{
    public static void main(String args[])
    {
        String a[] = new String[] {"Sam", "Claudia", "Josh", "Toby", "Donna"};
        int b[] = new int[] {1, 2, 3, 4, 5};

        for(int n=0;n<5;n++)
        {
            System.out.print (a[n] + "...");
            System.out.println (b[n]);
        }
        System.out.println (" ");

        java.util.Arrays.sort(a);

        for(int n=0;n<5;n++)
        {
            System.out.print (a[n] + "...");
            System.out.println (b[n]);
        }
    }

In a nutshell, this class created two arrays with five spaces each. It fills one with names of characters from the West Wing, and fills the other with numbering from one to five. We can say that the data in these two strings corresponds to each other.

Now, the program sorts the array with the names in it using Arrays.sort() . After printing the array again, you can see that while the names are now in alphabetical order, the numbers do not correspond anymore as the second array is unchanged.

How can I shuffle the contents of the second array to match the sort requirements of the first? The solution must also be flexible to allow for changes in the scope and size of the program. Please do not post any answers asking me to change my methodology with the arrays, or propose a more 'efficient' way of doing things. This is for educational purposed and I'd like a straight solution to the example code provided. Thanks in advance!

EDIT: I do NOT want to create an additional class, however I think some form of sorting through nested loops might be an option instead of Arrays.sort().

Answer 1

Below is the code without using any Map Collection, but if you want to use Map then it becomes very easy. Add both the arrays into map and sort it.

public static void main(String args[]) {
    String a[] = new String[] {
        "Sam", "Claudia", "Josh", "Toby", "Donna"
    };
    int b[] = new int[] {
        1, 2, 3, 4, 5
    };
    for (int n = 0; n < 5; n++) {
        System.out.print(a[n] + "...");
        System.out.println(b[n]);
    }
    System.out.println(" ");
    //java.util.Arrays.sort(a);
    /* Bubble Sort */
    for (int n = 0; n < 5; n++) {
        for (int m = 0; m < 4 - n; m++) {
            if ((a[m].compareTo(a[m + 1])) > 0) {
                String swapString = a[m];
                a[m] = a[m + 1];
                a[m + 1] = swapString;
                int swapInt = b[m];
                b[m] = b[m + 1];
                b[m + 1] = swapInt;
            }
        }
    }
    for (int n = 0; n < 5; n++) {
        System.out.print(a[n] + "...");
        System.out.println(b[n]);
    }
}

Answer 2

Some people propose making a product type. That is feasible only if the amount of elements is small. By introducing another object you add object overhead (30+ bytes) for each element and a performance penalty of a pointer (also worsening cache locality).

Solution without object overhead

Make a third array. Fill it with indices from 0 to size-1 . Sort this array with comparator function polling into the array according to which you want to sort.

Finally, reorder the elements in both arrays according to indices.

Alternative solution

Write the sorting algorithm yourself. This is not ideal, because you might make a mistake and the sorting efficiency might be subpar.

Answer 3

You have to ZIP your two arrays into an array which elements are instances of a class like:

class NameNumber 
{

    public NameNumber(String name, int n) {
        this.name = name;
        this.number = n;
    }

    public String name;
    public int number;
}  

And sort that array with a custom comparator .

Your code should be something like:

NameNumber [] zip = new NameNumber[Math.min(a.length,b.length)];
for(int i = 0; i < zip.length; i++)
{
    zip[i] = new NameNumber(a[i],b[i]);
}

Arrays.sort(zip, new Comparator<NameNumber>() {

    @Override
    public int compare(NameNumber o1, NameNumber o2) {
        return Integer.compare(o1.number, o2.number);
    }
});

Answer 4

You should not have two parallel arrays. Instead, you should have a single array of WestWingCharacter objects, where each object would have a field name and a field number .

Sorting this array by number of by name would then be a piece of cake:

Collections.sort(characters, new Comparator<WestWingCharacter>() {
    @Override
    public int compare(WestWingCharacter c1, WestWingCharacter c2) {
        return c1.getName().compareTo(c2.getName();
    }
});

or, with Java 8:

Collections.sort(characters, Comparator.comparing(WestWingCharacter::getName));

Java is an OO language, and you should thus use objects.

Answer 5

What you want is not possible because you don't know internally how Arrays.sort swap the elements in your String array, so there is no way to swap accordingly the elements in the int array.

You should create a class that contains the String name and the int position as parameter and then sort this class only with the name, providing a custom comparator to Arrays.sort .

If you want to keep your current code (with 2 arrays, but this not the ideal solution), don't use Arrays.sort and implement your own sorting algorithm. When you swap two names, get the index of them and swap the two integers in the other array accordingly.

Answer 6

Here is the answer for your query.

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

  String name[] = new String[] {"Sam", "Claudia", "Josh", "Toby", "Donna"};
        int id[] = new int[] {1, 2, 3, 4, 5};

        for ( int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int dtmp=0;
                String stmp=null;
                if (id[i] > id[j]) {
                    dtmp = rate[i];
                    id[i] = id[j];
                    id[j] = dtmp;
                    stmp = name[i];
                    name[i]=name[j];
                    name[j]=stmp;
                }
            }
        }
        System.out.println("Details are :");
        for(int i=0;i<n;i++){
            System.out.println(name[i]+" - "+id[i]);
        }
    }
}

Answer 7

The arrays are not linked in any way. Like someone pointed out take a look at

SortedMap http://docs.oracle.com/javase/7/docs/api/java/util/SortedMap.html

TreeMaphttp://docs.oracle.com/javase/7/docs/api/java/util/TreeMap.html

Answer 8

   import java.util.*;

class mergeArrays2
{
  public static void main(String args[])
  {
     String a1[]={"Sam", "Claudia", "Josh", "Toby", "Donna"};
     Integer a2[]={11, 2, 31, 24, 5};

     ArrayList ar1=new ArrayList(Arrays.asList(a1));

     Collections.sort(ar1);

     ArrayList ar2=new ArrayList(Arrays.asList(a2));

     Collections.sort(ar2);





     System.out.println("array list"+ar1+ " "+ar2);
  }

}