如何从hashmap中获取5个最高值？

Question

I have a Hashmap that links a zipcodes stored as keys and population stored as values in a hashmap.

The hashmap contains around 33k entries.

I'm trying to get the 5 highest population values from 5 zip codes and print out the 5 zip codes ASSOCIATED with the 5 highest population, but I'm having trouble understanding the algorithm of how to do it.

If it was just one, its easy but the 5 restriction is giving me some trouble.

I know to store the 5 values in an int array and I have a counter to determine when 5 of them are stored, but thats it.

Thanks

    int populatedCounter = 0;

    int[] populatedZip = new int[5];

    it = zipCodePop.entrySet().iterator();
    while (it.hasNext())
    {
        Map.Entry pairs = (Map.Entry)it.next();

        for (int i = 0; i < populatedZip.length; i++)
        {

        }
    }

}

Answer 1

Putting the entries of such a set into a list and sorting it is one option. But 33k elements is a number where the O(n*log(n)) complexity of sorting might already have a noticable performance impact.

One apporach would be to employ the PriorityQueue that nr4bt already mentioned (I wrote this snippet while he answered). It basically inserts all elements into a PriorityQueue that is sorted according to the values of the map entries.

import java.util.ArrayList;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.PriorityQueue;

public class GreatestOfMap
{
    public static void main(String[] args)
    {
        Map<String, Integer> map = new HashMap<String, Integer>();

        map.put("zip000", 1234);
        map.put("zip001", 2345);
        map.put("zip002", 3456);
        map.put("zip003", 4567);
        map.put("zip004", 5678);
        map.put("zip005", 6789);
        map.put("zip006", 123);
        map.put("zip007", 234);
        map.put("zip008", 456);
        map.put("zip009", 567);
        map.put("zip010", 7890);
        map.put("zip011", 678);
        map.put("zip012", 789);
        map.put("zip013", 890);

        int n = 5;
        List<Entry<String, Integer>> greatest = findGreatest(map, 5);
        System.out.println("Top "+n+" entries:");
        for (Entry<String, Integer> entry : greatest)
        {
            System.out.println(entry);
        }
    }

    private static <K, V extends Comparable<? super V>> List<Entry<K, V>> 
        findGreatest(Map<K, V> map, int n)
    {
        Comparator<? super Entry<K, V>> comparator = 
            new Comparator<Entry<K, V>>()
        {
            @Override
            public int compare(Entry<K, V> e0, Entry<K, V> e1)
            {
                V v0 = e0.getValue();
                V v1 = e1.getValue();
                return v0.compareTo(v1);
            }
        };
        PriorityQueue<Entry<K, V>> highest = 
            new PriorityQueue<Entry<K,V>>(n, comparator);
        for (Entry<K, V> entry : map.entrySet())
        {
            highest.offer(entry);
            while (highest.size() > n)
            {
                highest.poll();
            }
        }

        List<Entry<K, V>> result = new ArrayList<Map.Entry<K,V>>();
        while (highest.size() > 0)
        {
            result.add(highest.poll());
        }
        return result;
    }
}

Answer 2

Try this, using standard methods and assuming that the population count is stored as Integer s in the HashMap :

List<Integer> list = new ArrayList<Integer>(zipCodePop.values());
Collections.sort(list, Collections.reverseOrder());
List<Integer> top5 = list.subList(0, 5);

Answer 3

public class CheckHighiestValue { public static void main(String... s) {

    HashMap<String, Integer> map = new HashMap<String, Integer>();

    map.put("first", 10000);
    map.put("second", 20000);
    map.put("third", 300);
    map.put("fourth", 800012);
    map.put("fifth", 5000);
    map.put("sixth", 30012);
    map.put("seventh", 1234);
    map.put("eighth", 45321);
    map.put("nineth", 5678);

    Set<Entry<String, Integer>> set = map.entrySet();

    List<Entry<String, Integer>> list = new ArrayList<Entry<String, Integer>>(
            set);

    Collections.sort(list, new Comparator<Map.Entry<String, Integer>>() {

        @Override
        public int compare(Entry<String, Integer> o1,
                Entry<String, Integer> o2) {

            return o2.getValue().compareTo(o1.getValue());
        }

    });
    System.out.println(list.subList(0, 5));
}

}

Answer 4

PriorityQueue would help too, and also a nice topic about how to get top k from a list, you can check this link

PriorityQueue<Integer> p = new PriorityQueue<Integer>(5);

int[] a = new int[]{3,5,10,1,23,42,66,1333,545,110};

for (int i : a){
    p.add(i);
    if (p.size() > 5){
        p.poll();
    }
}

//output will be highest 5, [42, 66, 110, 1333, 545]

You can have O(n log(k)) time complexity // k is your top value count.

Answer 5

This is something i made and hopefully provides you something that you want to use.

public class TopsCollection { 

private static Map<String, Integer> collectors = new HashMap<>();

public TopsCollection() {
}

public void add(String playerName, int score) {
    collectors.put(playerName, score);
}

public void clearCollectors() {
    synchronized (collectors) {
        collectors.clear();
    }
}

public List<Map.Entry<String, Integer>> getTops() {
    return collectors.entrySet().stream().sorted(comparing(Map.Entry::getValue, reverseOrder())).limit(5).collect(toList());
}

public int getTopByName(String name) {
    for (int i = 0; i < getTops().size(); i++) {
        if (getTops().get(i).getKey().contains(name)) {
            return i;
        }
    }
    return 0;
}

getTopByName allows you to get the top place of the specified name.

Answer 6

How would you do this without a computer, with just a piece of paper and a pencil? Pretend you had a stack of index cards that had numbers on them, and it was your job to find the 5 highest numbers. How would you do that? Write down steps that somebody else could follow to achieve the goal, and when you have those steps written out, you'll have an algorithm that you can start thinking about implementing with code.

You say that a single maximum is easy, so do it exactly like you would with a single maximum, but keep track of the five maximums instead. An array of maximums might be helpful here.