哈希表中最大值的键

Question

嗨，我有以下对象：

Hashtable<Object, Double>

我想在表中找到最大双精度值的键。最简单的方法是什么？

谢谢

Answer 1

没有内置的函数可以从Hashtable中获取最大值，您必须遍历所有键并手动确定最大值。

Object maxKey=null;
Double maxValue = Double.MIN_VALUE; 
for(Map.Entry<Object,Double> entry : table.entrySet()) {
     if(entry.getValue() > maxValue) {
         maxValue = entry.getValue();
         maxKey = entry.getKey();
     }
}

编辑:查找超过1个键的最大值

ArrayList<Object> maxKeys= new ArrayList<Object>();
Double maxValue = Double.MIN_VALUE; 
for(Map.Entry<Object,Double> entry : table.entrySet()) {
     if(entry.getValue() > maxValue) {
         maxKeys.clear(); /* New max remove all current keys */
         maxKeys.add(entry.getKey());
         maxValue = entry.getValue();
     }
     else if(entry.getValue() == maxValue)
     {
       maxKeys.add(entry.getKey());
     }
}

Answer 2

如果它真的很重要，您可以在不迭代所有键的情况下完成，只需扩展HashTable

class MyHashtable extends Hashtable<Object, Double> {

    private Double maxValue = Double.MIN_VALUE;

    @Override
    public synchronized Double put(Object k, Double v) {
        maxValue = Math.max(maxValue, v);
        return super.put(k, v);
    }

    @Override
    public synchronized void clear() {
        super.clear();
        maxValue = Double.MIN_VALUE;
    }

    public Double getMaxValue() {
        return maxValue;
    }

    @Override
    public synchronized Double remove(Object key) {
        // TODO: Left as an Excercise for the user, refer the other answers
        return super.remove(key);
    }
}

Answer 3

您可以遍历并找到最大值：

public static void main(String[] args) {
    Map<Object, Double> maps = new HashMap<Object, Double>();
    maps.put("5", new Double(50.0));
    maps.put("4", new Double(40.0));
    maps.put("2", new Double(20.0));
    maps.put("1", new Double(100.0));
    maps.put("3", new Double(30.0));
    maps.put("5", new Double(50.0));

    Double max = Double.MIN_VALUE;
    for(Object key: maps.keySet()) {
        Double tmp = maps.get(key);
        if(tmp.compareTo(max) > 0) {
            max = tmp;
        }
    }

    System.out.println(max);
}