从列表中查找最少 3 个值

Question

I am working on project where i need to find the minimum 3 three values from a HashMap . I have found a code that finds highest values. Here it is.

public 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>>()
    {
        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 (java.util.Map.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;
    }

How cam I modify it to find the minimum values? Kindly help me here. UPDATE :Sorry for my mistake, I was looking to find minimum values from a HashMap

Answer 1

就个人而言，我会简单地通过Collections.sort( ... ) （因为它似乎由 Comparables 组成Collections.sort( ... )对列表进行排序，然后返回三个第一个（或最后一个）元素。

Answer 2

You could use exactly the same code except reverse the order of the Comparator using Collections.reverseOrder(Comparator c) .

PriorityQueue<Entry<K, V>> highest = 
    new PriorityQueue<Entry<K,V>>(n, Collections.reverseOrder(comparator));

However, that would not be the most efficient approach. It is possible to do this in one pass.

Answer 3

Simply you can try like this. You don't need to have a customized comparator to achieve what you are trying here.

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

    list.add(23);
    list.add(65);
    list.add(85);
    list.add(1);
    list.add(54);
    list.add(5);

    Collections.sort(list);

    for(int i=0;i<3;i++)
        System.out.println(list.get(i));

Answer 4

您可以使用Collections.min(list)从列表中获取最低值

Answer 5

The suggestions involving sorting the map have a runtime of O(n log(n)). The following implementation has a runtime of O(nm), where n is the size of the map and m is the size of the result, ie if you want the m smallest elements.

public static <K, V extends Comparable<? super V>> List<Entry<K, V>> findGreatest(Map<K, V> map, int m)
{
    LinkedList<Entry<K, V>> result = new LinkedList<Entry<K, V>>();

    for (Entry<K, V> entry : map.entrySet())
    {
        ListIterator<Entry<K, V>> iterator = result.listIterator();
        int i = n;
        while (iterator.hasNext() && i-- > 0)
        {
            if (iterator.next().getValue().compareTo(entry.getValue()) > 0)
            {
                iterator.previous();
                iterator.add(entry);
                if (result.size() > n)
                {
                    result.removeLast();
                }
                break;
            }
        }

        if (result.size() < n)
        {
            result.add(entry);
        }
    }

    return result;
}

Answer 6

@Jude Niroshan, his answer also correct one and you can try this also if it helps, you can change the values and check.

LinkedList<Integer> li=new LinkedList<Integer>();
    li.add(-8);
    li.add(0);
    li.add(10);
    li.add(-1);
    li.add(9);
Collections.sort(li);
for(Integer a:li){
        if(a<3)
        System.out.println(a);
    }