根据另一个数组对数组排序

Question

I have 2 arrays :

The first one consists of names of persons . The second one gives me a count of number of names that have appeared in a page. Ex: Names=["James","Loiui","Mario","Peter"] count=[1,4,2,5]

now I tried with this code give below :

SortedMap<Integer, String> m = new TreeMap<Integer, String>();
for(int i = 0; i < Names.size(); i++)
        m.put( count.get(i),Names.get(i));

But this doesnt work given there are repeating values of count . I figured out that the problem is with TreeMap as it stores only unique elements . Now to overcome problem of mine, are there any other valid solutions .

Answer 1

Create a class PersonFrequency , containing a name and a count. Create a single array or list of PersonFrequency instances. Sort this array by count:

Java 8 example:

List<PersonFrequency> list = new ArrayList<>(names.length);
for (int i = 0; i < names.length; i++) {
    list.add(new PersonFrequency(names[i], counts[i]);
}

list.sort(Comparator.comparing(PersonFrequency::getCount).reversed());

List<String> sortedNames = list.stream()
                               .map(PersonFrequency::getName)
                               .collect(Collectors.toList());

In Java 7, the sort would become

Collections.sort(list, new Comparator<PersonFrequency>() {
    @Override
    public int compare(PersonFrequency p1, PersonFrequency p2) {
        return Integer.compare(p2.getCount(), p1.getCount());
    }
});

Answer 2

Sorted alphabetically

    String [] names = {"Aaa","Bbb"};
    int [] count = {1,2};

    SortedMap<String, Integer> m = new TreeMap<String, Integer>();

    for(int i = 0; i < names.length; i++){
        m.put(names[i], count[i]);
    }

Sorted by number:

    String [] names = {"Aaa","Bbb"};
    int [] count = {2,1};

    SortedMap<Integer, String> m = new TreeMap<Integer, String>();

    for(int i = 0; i < names.length; i++){
        m.put(count[i],names[i]);
    }

For descending change your sorted map to this

SortedMap<Integer, String> m = new TreeMap<Integer, String>().descendingMap();

Answer 3

As mentioned in the previous comments you can have a sorted map with Integer and ArrayList parameters as below,

SortedMap<Integer, ArrayList<String>> map=new TreeMap<Integer, ArrayList<String>>().descendingMap();

Since you want the results to be in descending order I added descendingMap().

We use ArrayList because you may have duplicates in count array. More than one name can have the same count value. So one count will have a list of names.

And the complete code is below,

import java.util.ArrayList;
import java.util.SortedMap;
import java.util.TreeMap;



public class NameCount {

    public static void main(String[] args) {

        String[] names = {"James","Loiui","Mario","Peter"};
        int[] count = {1, 4, 2, 1};

        SortedMap<Integer, ArrayList<String>> map=new TreeMap<Integer, ArrayList<String>>().descendingMap();
        ArrayList<String> nameList=null;

        for(int i = 0; i < names.length; i++)
        {
            if(map.get(count[i])==null)
            {
                nameList = new ArrayList<String>();
                nameList.add(names[i]);
                map.put(count[i], nameList);
            }
            else
            {
                map.get(count[i]).add(names[i]);
            }
        }

        for(int countVal : map.keySet())
        {
            ArrayList<String> namesListVal=map.get(countVal);

            System.out.print("\nCount  "+countVal+ ":  NAMES  : ");
            for(String name : namesListVal)
            {
                System.out.print(name+"  ");
            }

        }
    }

}