Java:是不是有一个容器可以有效地结合 HashMap 和 ArrayList?
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【中文标题】Java:是不是有一个容器可以有效地结合 HashMap 和 ArrayList?【英文标题】:Java: Is there a container which effectively combines HashMap and ArrayList?Java:是否有一个容器可以有效地结合 HashMap 和 ArrayList? 【发布时间】:2011-07-08 17:46:48 【问题描述】:我不断发现需要一个容器,它既是HashMap
(用于快速查找键类型)又是ArrayList
(用于通过整数索引快速访问)。
LinkedHashMap
几乎是正确的,因为它保留了一个可迭代列表,但不幸的是它是一个链接 列表...检索第 N 个元素需要从 1 迭代到 N。
有没有一种容器类型符合这个要求,但我不知何故错过了它?当其他人需要通过键和通过索引访问同一组数据时,他们会怎么做?
【问题讨论】:
您可以将所有对象同时添加到 HashMap 和 ArrayList 中吗? @Blorgbeard,那么当您从地图中删除时会发生什么,从 ArrayList 的中间删除?你能得到的最好的就是 LinkedHashMap 之类的(这是我在 java.util 中最喜欢的数据结构) 到目前为止,我想不出有一次我必须使用索引访问 LinkedHashMap,可能算法/数据结构可以重新设计。例如,如果您尝试基于某个行索引访问 Map,您最好将键也保留在该行中,这样您就可以使用键而不是索引来检索数据。 @bestsss - 是的,你必须从 ArrayList 中删除。因此,您在删除(或插入)时会受到 O(n) 的影响,但在按索引获取时会受到 O(1) 的影响——而 LinkedHashMap 则相反,对吧? @Blorgbeard,是的,当然。对于大 N 来说,它可能会使插入/删除相当重 【参考方案1】:看看 Apache Commons LinkedMap。
【讨论】:
这正是我认为必须以标准形式存在的东西。它有最重要的 get(int index) 成员。非常感谢! 按名称,我仍然认为get(index)
查找具有 O(index)(或 O(min(index, n-index))
)复杂性。
@Paŭlo,是的,你是对的,它是基于链表实现的。你可以像 LinkedHashMap 一样做。
@Reuben,对于像 get(index) 这样的操作,没有办法实现获得 O(1) 的哈希映射。【参考方案2】:
如果您要删除(在中间)以及通过索引和键访问(这意味着索引正在更改),您可能会看不出来 - 我认为根本不可能有提供O(1)
用于remove
(按索引、键或迭代器)和get(index)
。这就是为什么我们在标准 API 中同时拥有 LinkedList
(O(1)
中的 iterator.remove()
或 remove(0)
)和 ArrayList(O(1)
中的 get(index)
)。
如果您使用树结构而不是数组或链表(可以与基于 O(1) 键的读取访问相结合 - 获取您的键的索引,则您可以在 O(log n)
中同时进行删除和索引获取不过,-value-pair 仍然需要 O(log n)。
如果您不想删除任何内容,或者可以忍受以下索引未移动(即remove(i)
等同于set(i, null)
,则没有任何内容禁止同时拥有 O(1) 索引和键访问 - 在事实上,这里的索引只是第二个键,所以你可以简单地使用一个 HashMap 和一个 ArrayList(或两个 HashMap),然后用一个瘦包装器将两者结合起来。
编辑:因此,这里是ArrayHashMap
的实现,如上面最后一段所述(使用“昂贵的删除”变体)。它实现了下面的接口IndexedMap
。 (如果你不想在这里复制+粘贴,两者也在我的github account 中,以防以后更改)。
package de.fencing_game.paul.examples;
import java.util.*;
/**
* A combination of ArrayList and HashMap which allows O(1) for read and
* modifiying access by index and by key.
* <p>
* Removal (either by key or by index) is O(n), though,
* as is indexed addition of a new Entry somewhere else than the end.
* (Adding at the end is in amortized O(1).)
* </p>
* <p>
* (The O(1) complexity for key based operations is under the condition
* "if the hashCode() method of the keys has a suitable distribution and
* takes constant time", as for any hash-based data structure.)
* </p>
* <p>
* This map allows null keys and values, but clients should think about
* avoiding using these, since some methods return null to show
* "no such mapping".
* </p>
* <p>
* This class is not thread-safe (like ArrayList and HashMap themselves).
* </p>
* <p>
* This class is inspired by the question
* <a href="http://***.com/questions/5192706/java-is-there-a-container-which-effectively-combines-hashmap-and-arraylist">Is there a container which effectively combines HashMap and ArrayList?</a> on ***.
* </p>
* @author Paŭlo Ebermann
*/
public class ArrayHashMap<K,V>
extends AbstractMap<K,V>
implements IndexedMap<K,V>
/**
* Our backing map.
*/
private Map<K, SimpleEntry<K,V>> baseMap;
/**
* our backing list.
*/
private List<SimpleEntry<K,V>> entries;
/**
* creates a new ArrayHashMap with default parameters.
* (TODO: add more constructors which allow tuning.)
*/
public ArrayHashMap()
this.baseMap = new HashMap<K,SimpleEntry<K,V>>();
this.entries = new ArrayList<SimpleEntry<K,V>>();
/**
* puts a new key-value mapping, or changes an existing one.
*
* If new, the mapping gets an index at the end (i.e. @link #size()
* before it gets increased).
*
* This method runs in O(1) time for changing an existing value,
* amortized O(1) time for adding a new value.
*
* @return the old value, if such, else null.
*/
public V put(K key, V value)
SimpleEntry<K,V> entry = baseMap.get(key);
if(entry == null)
entry = new SimpleEntry<K,V>(key, value);
baseMap.put(key, entry);
entries.add(entry);
return null;
return entry.setValue(value);
/**
* retrieves the value for a key.
*
* This method runs in O(1) time.
*
* @return null if there is no such mapping,
* else the value for the key.
*/
public V get(Object key)
SimpleEntry<K,V> entry = baseMap.get(key);
return entry == null ? null : entry.getValue();
/**
* returns true if the given key is in the map.
*
* This method runs in O(1) time.
*
*/
public boolean containsKey(Object key)
return baseMap.containsKey(key);
/**
* removes a key from the map.
*
* This method runs in O(n) time, n being the size of this map.
*
* @return the old value, if any.
*/
public V remove(Object key)
SimpleEntry<K,V> entry = baseMap.remove(key);
if(entry == null)
return null;
entries.remove(entry);
return entry.getValue();
/**
* returns a key by index.
*
* This method runs in O(1) time.
*
*/
public K getKey(int index)
return entries.get(index).getKey();
/**
* returns a value by index.
*
* This method runs in O(1) time.
*
*/
public V getValue(int index)
return entries.get(index).getValue();
/**
* Returns a set view of the keys of this map.
*
* This set view is ordered by the indexes.
*
* It supports removal by key or iterator in O(n) time.
* Containment check runs in O(1).
*/
public Set<K> keySet()
return new AbstractSet<K>()
public void clear()
entryList().clear();
public int size()
return entries.size();
public Iterator<K> iterator()
return keyList().iterator();
public boolean remove(Object key)
return keyList().remove(key);
public boolean contains(Object key)
return keyList().contains(key);
;
// keySet()
/**
* Returns a set view of the entries of this map.
*
* This set view is ordered by the indexes.
*
* It supports removal by entry or iterator in O(n) time.
*
* It supports adding new entries at the end, if the key
* is not already used in this map, in amortized O(1) time.
*
* Containment check runs in O(1).
*/
public Set<Map.Entry<K,V>> entrySet()
return new AbstractSet<Map.Entry<K,V>>()
public void clear()
entryList().clear();
public int size()
return entries.size();
public Iterator<Map.Entry<K,V>> iterator()
return entryList().iterator();
public boolean add(Map.Entry<K,V> e)
return entryList().add(e);
public boolean contains(Object o)
return entryList().contains(o);
public boolean remove(Object o)
return entryList().remove(o);
;
// entrySet()
/**
* Returns a list view of the entries of this map.
*
* This list view is ordered by the indexes.
*
* It supports removal by entry, iterator or sublist.clear in O(n) time.
* (n being the length of the total list, not the sublist).
*
* It supports adding new entries at the end, if the key
* is not already used in this map, in amortized O(1) time.
*
* Containment check runs in O(1).
*/
public List<Map.Entry<K,V>> entryList()
return new AbstractList<Map.Entry<K,V>>()
public void clear()
baseMap.clear();
entries.clear();
public Map.Entry<K,V> get(int index)
return entries.get(index);
public int size()
return entries.size();
public Map.Entry<K,V> remove(int index)
Map.Entry<K,V> e = entries.remove(index);
baseMap.remove(e.getKey());
return e;
public void add(int index, Map.Entry<K,V> newEntry)
K key = newEntry.getKey();
SimpleEntry<K,V> clone = new SimpleEntry<K,V>(newEntry);
if(baseMap.containsKey(key))
throw new IllegalArgumentException("duplicate key " +
key);
entries.add(index, clone);
baseMap.put(key, clone);
public boolean contains(Object o)
if(o instanceof Map.Entry)
SimpleEntry<K,V> inMap =
baseMap.get(((Map.Entry<?,?>)o).getKey());
return inMap != null &&
inMap.equals(o);
return false;
public boolean remove(Object o)
if (!(o instanceof Map.Entry))
Map.Entry<?,?> e = (Map.Entry<?,?>)o;
SimpleEntry<K,V> inMap = baseMap.get(e.getKey());
if(inMap != null && inMap.equals(e))
entries.remove(inMap);
baseMap.remove(inMap.getKey());
return true;
return false;
protected void removeRange(int fromIndex, int toIndex)
List<SimpleEntry<K,V>> subList =
entries.subList(fromIndex, toIndex);
for(SimpleEntry<K,V> entry : subList)
baseMap.remove(entry.getKey());
subList.clear();
;
// entryList()
/**
* Returns a List view of the keys in this map.
*
* It allows index read access and key containment check in O(1).
* Changing a key is not allowed.
*
* Removal by key, index, iterator or sublist.clear runs in O(n) time
* (this removes the corresponding values, too).
*/
public List<K> keyList()
return new AbstractList<K>()
public void clear()
entryList().clear();
public K get(int index)
return entries.get(index).getKey();
public int size()
return entries.size();
public K remove(int index)
Map.Entry<K,V> e = entries.remove(index);
baseMap.remove(e.getKey());
return e.getKey();
public boolean remove(Object key)
SimpleEntry<K,V> entry = baseMap.remove(key);
if(entry == null)
return false;
entries.remove(entry);
return true;
public boolean contains(Object key)
return baseMap.containsKey(key);
protected void removeRange(int fromIndex, int toIndex)
entryList().subList(fromIndex, toIndex).clear();
;
// keyList()
/**
* Returns a List view of the values in this map.
*
* It allows get and set by index in O(1) time (set changes the mapping).
*
* Removal by value, index, iterator or sublist.clear is possible
* in O(n) time, this removes the corresponding keys too (only the first
* key with this value for remove(value)).
*
* Containment check needs an iteration, thus O(n) time.
*/
public List<V> values()
return new AbstractList<V>()
public int size()
return entries.size();
public void clear()
entryList().clear();
public V get(int index)
return entries.get(index).getValue();
public V set(int index, V newValue)
Map.Entry<K,V> e = entries.get(index);
return e.setValue(newValue);
public V remove(int index)
Map.Entry<K,V> e = entries.remove(index);
baseMap.remove(e.getKey());
return e.getValue();
protected void removeRange(int fromIndex, int toIndex)
entryList().subList(fromIndex, toIndex).clear();
;
// values()
/**
* an usage example method.
*/
public static void main(String[] args)
IndexedMap<String,String> imap = new ArrayHashMap<String, String>();
for(int i = 0; i < args.length-1; i+=2)
imap.put(args[i], args[i+1]);
System.out.println(imap.values());
System.out.println(imap.keyList());
System.out.println(imap.entryList());
System.out.println(imap);
System.out.println(imap.getKey(0));
System.out.println(imap.getValue(0));
界面如下:
package de.fencing_game.paul.examples;
import java.util.*;
/**
* A map which additionally to key-based access allows index-based access
* to keys and values.
* <p>
* Inspired by the question <a href="http://***.com/questions/5192706/java-is-there-a-container-which-effectively-combines-hashmap-and-arraylist">Is there a container which effectively combines HashMap and ArrayList?</a> on ***.
* </p>
* @author Paŭlo Ebermann
* @see ArrayHashMap
*/
public interface IndexedMap<K,V>
extends Map<K,V>
/**
* returns a list view of the @link #entrySet of this Map.
*
* This list view supports removal of entries, if the map is mutable.
*
* It may also support indexed addition of new entries per the
* @link List#add add method - but this throws an
* @link IllegalArgumentException if the key is already used.
*/
public List<Map.Entry<K,V>> entryList();
/**
* returns a list view of the @link #keySet.
*
* This list view supports removal of keys (with the corresponding
* values), but does not support addition of new keys.
*/
public List<K> keyList();
/**
* returns a list view of values contained in this map.
*
* This list view supports removal of values (with the corresponding
* keys), but does not support addition of new values.
* It may support the @link List#set set operation to change the
* value for a key.
*/
public List<V> values();
/**
* Returns a value of this map by index.
*
* This is equivalent to
* @ #values() values().@link List#get get@code (index).
*/
public V getValue(int index);
/**
* Returns a key of this map by index.
*
* This is equivalent to
* @ #keyList keyList().@link List#get get@code (index).
*/
public K getKey(int index);
【讨论】:
ŭlo,你知道 LinkedList 的 remove(index) O(1) 有什么好玩的,找到索引本身是 O(n) 遍历链接结构(这比 ArrayList 变体慢) 对于迭代器删除它是 O(1),不过。是的,也应该这样写。【参考方案3】:你可以自己做,但这里是example implemenation。相应的谷歌搜索词将是“ArrayMap”。
我不确定,但也许 commons 收藏或 google 收藏有这样的地图。
编辑:
您可以创建一个使用 arraylist 实现的 hashmap,即它可以像 LinkedHashMap 一样工作,因为插入顺序定义了列表索引。这将提供快速 get(Index) (O(1)) 和 get(Name) (O(1)) 访问,插入也将是 O(1)(除非必须扩展数组),但删除将是 O (n) 因为删除第一个元素需要更新所有索引。
这个技巧可以通过一个内部保存一个 Map for key-> index 然后是一个 ArrayList 的映射来完成。
get(Key) 将是(没有错误检查的简单示例):
list.get(keyIndexMap.get(key));
【讨论】:
这有一个 O(1) 索引访问(如果我们要添加一个 get(index) 方法),但get(key)
有一个 O(n)
复杂度。【参考方案4】:
您为什么不简单地保留HashMap
,然后按照建议的here 使用hashMap.entrySet().toArray();
?
【讨论】:
这不是复制吗?我一直认为 toArray() 非常昂贵(至少对于大列表)... 大约是O(n+capacity)
- 所以你不会每次都这样做,只有在填满地图之后。它不适合经常变化的地图。以上是关于Java:是不是有一个容器可以有效地结合 HashMap 和 ArrayList?的主要内容,如果未能解决你的问题,请参考以下文章