我认为你的选择是

Use a map implementation built for this, like the BiMap from google collections. Note that the google collections BiMap requires uniqueless of values, as well as keys, but it provides high performance in both directions performance Manually maintain two maps - one for key -> value, and another map for value -> key Iterate through the entrySet() and to find the keys which match the value. This is the slowest method, since it requires iterating through the entire collection, while the other two methods don't require that.

使用Java 8:

ftw.forEach((key, value) -> {
    if (value.equals("foo")) {
        System.out.print(key);
    }
});

使用薄包装:HMap

import java.util.Collections;
import java.util.HashMap;
import java.util.Map;

public class HMap<K, V> {

   private final Map<K, Map<K, V>> map;

   public HMap() {
      map = new HashMap<K, Map<K, V>>();
   }

   public HMap(final int initialCapacity) {
      map = new HashMap<K, Map<K, V>>(initialCapacity);
   }

   public boolean containsKey(final Object key) {
      return map.containsKey(key);
   }

   public V get(final Object key) {
      final Map<K, V> entry = map.get(key);
      if (entry != null)
         return entry.values().iterator().next();
      return null;
   }

   public K getKey(final Object key) {
      final Map<K, V> entry = map.get(key);
      if (entry != null)
         return entry.keySet().iterator().next();
      return null;
   }

   public V put(final K key, final V value) {
      final Map<K, V> entry = map
            .put(key, Collections.singletonMap(key, value));
      if (entry != null)
         return entry.values().iterator().next();
      return null;
   }
}

据我所知，当你将HashMap的键和值表示为数组时，它们是不混合的:

hashmap.values().toArray()

and

hashmap.keySet().toArray()

所以下面的代码(从java 8开始)应该像预期的那样工作:

public Object getKeyByFirstValue(Object value) {
    int keyNumber =  Arrays.asList(hashmap.values().toArray()).indexOf(value);
    return hashmap.keySet().toArray()[keyNumber];
}

然而，(警告!)它的工作速度比迭代慢2-3倍。

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Set;

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

        HashMap<String, List<String>> resultHashMap = new HashMap<String, List<String>>();

        Set<String> newKeyList = resultHashMap.keySet();


        for (Iterator<String> iterator = originalHashMap.keySet().iterator(); iterator.hasNext();) {
            String hashKey = (String) iterator.next();

            if (!newKeyList.contains(originalHashMap.get(hashKey))) {
                List<String> loArrayList = new ArrayList<String>();
                loArrayList.add(hashKey);
                resultHashMap.put(originalHashMap.get(hashKey), loArrayList);
            } else {
                List<String> loArrayList = resultHashMap.get(originalHashMap
                        .get(hashKey));
                loArrayList.add(hashKey);
                resultHashMap.put(originalHashMap.get(hashKey), loArrayList);
            }
        }

        System.out.println("Original HashMap : " + originalHashMap);
        System.out.println("Result HashMap : " + resultHashMap);
    }
}

如果你的数据结构在键和值之间有多对一映射，你应该遍历条目并选择所有合适的键:

public static <T, E> Set<T> getKeysByValue(Map<T, E> map, E value) {
    Set<T> keys = new HashSet<T>();
    for (Entry<T, E> entry : map.entrySet()) {
        if (Objects.equals(value, entry.getValue())) {
            keys.add(entry.getKey());
        }
    }
    return keys;
}

如果是一对一的关系，你可以返回第一个匹配的键:

public static <T, E> T getKeyByValue(Map<T, E> map, E value) {
    for (Entry<T, E> entry : map.entrySet()) {
        if (Objects.equals(value, entry.getValue())) {
            return entry.getKey();
        }
    }
    return null;
}

在Java 8中:

public static <T, E> Set<T> getKeysByValue(Map<T, E> map, E value) {
    return map.entrySet()
              .stream()
              .filter(entry -> Objects.equals(entry.getValue(), value))
              .map(Map.Entry::getKey)
              .collect(Collectors.toSet());
}

此外，对于番石榴用户来说，BiMap可能也很有用。例如:

BiMap<Token, Character> tokenToChar = 
    ImmutableBiMap.of(Token.LEFT_BRACKET, '[', Token.LEFT_PARENTHESIS, '(');
Token token = tokenToChar.inverse().get('(');
Character c = tokenToChar.get(token);