如果你的關鍵字符串有特定的模式，特別是常見的根，那麼一個Trie可能是一個存儲少得多的數據的有效方法。
下面是工作TrieMap的代碼。
注：使用EntrySet跨越Map s到迭代不不適用於Trie S中的常用建議。它們在Trie中效率非常低，所以請儘可能避免請求。
/**
* Implementation of a Trie structure.
*
* A Trie is a compact form of tree that takes advantage of common prefixes
* to the keys.
*
* A normal HashSet will take the key and compute a hash from it, this hash will
* be used to locate the value through various methods but usually some kind
* of bucket system is used. The memory footprint resulting becomes something
* like O(n).
*
* A Trie structure essentuially combines all common prefixes into a single key.
* For example, holding the strings A, AB, ABC and ABCD will only take enough
* space to record the presence of ABCD. The presence of the others will be
* recorded as flags within the record of ABCD structure at zero cost.
*
* This structure is useful for holding similar strings such as product IDs or
* credit card numbers.
*
*/
public class TrieMap<V> extends AbstractMap<String, V> implements Map<String, V> {
/**
* Map each character to a sub-trie.
*
* Could replace this with a 256 entry array of Tries but this will handle
* multibyte character sets and I can discard empty maps.
*
* Maintained at null until needed (for better memory footprint).
*
*/
protected Map<Character, TrieMap<V>> children = null;
/**
* Here we store the map contents.
*/
protected V leaf = null;
/**
* Set the leaf value to a new setting and return the old one.
*
* @param newValue
* @return old value of leaf.
*/
protected V setLeaf(V newValue) {
V old = leaf;
leaf = newValue;
return old;
}
/**
* I've always wanted to name a method something like this.
*/
protected void makeChildren() {
if (children == null) {
// Use a TreeMap to ensure sorted iteration.
children = new TreeMap<Character, TrieMap<V>>();
}
}
/**
* Finds the TrieMap that "should" contain the key.
*
* @param key
*
* The key to find.
*
* @param grow
*
* Set to true to grow the Trie to fit the key.
*
* @return
*
* The sub Trie that "should" contain the key or null if key was not found and
* grow was false.
*/
protected TrieMap<V> find(String key, boolean grow) {
if (key.length() == 0) {
// Found it!
return this;
} else {
// Not at end of string.
if (grow) {
// Grow the tree.
makeChildren();
}
if (children != null) {
// Ask the kids.
char ch = key.charAt(0);
TrieMap<V> child = children.get(ch);
if (child == null && grow) {
// Make the child.
child = new TrieMap<V>();
// Store the child.
children.put(ch, child);
}
if (child != null) {
// Find it in the child.
return child.find(tail(key), grow);
}
}
}
return null;
}
/**
* Remove the head (first character) from the string.
*
* @param s
*
* The string.
*
* @return
*
* The same string without the first (head) character.
*
*/
// Suppress warnings over taking a subsequence
private String tail(String s) {
return s.substring(1, s.length());
}
/**
*
* Add a new value to the map.
*
* Time footprint = O(s.length).
*
* @param s
*
* The key defining the place to add.
*
* @param value
*
* The value to add there.
*
* @return
*
* The value that was there, or null if it wasn't.
*
*/
@Override
public V put(String key, V value) {
V old = null;
// If empty string.
if (key.length() == 0) {
old = setLeaf(value);
} else {
// Find it.
old = find(key, true).put("", value);
}
return old;
}
/**
* Gets the value at the specified key position.
*
* @param o
*
* The key to the location.
*
* @return
*
* The value at that location, or null if there is no value at that location.
*/
@Override
public V get(Object o) {
V got = null;
if (o != null) {
String key = (String) o;
TrieMap<V> it = find(key, false);
if (it != null) {
got = it.leaf;
}
} else {
throw new NullPointerException("Nulls not allowed.");
}
return got;
}
/**
* Remove the value at the specified location.
*
* @param o
*
* The key to the location.
*
* @return
*
* The value that was removed, or null if there was no value at that location.
*/
@Override
public V remove(Object o) {
V old = null;
if (o != null) {
String key = (String) o;
if (key.length() == 0) {
// Its me!
old = leaf;
leaf = null;
} else {
TrieMap<V> it = find(key, false);
if (it != null) {
old = it.remove("");
}
}
} else {
throw new NullPointerException("Nulls not allowed.");
}
return old;
}
/**
* Count the number of values in the structure.
*
* @return
*
* The number of values in the structure.
*/
@Override
public int size() {
// If I am a leaf then size increases by 1.
int size = leaf != null ? 1 : 0;
if (children != null) {
// Add sizes of all my children.
for (Character c : children.keySet()) {
size += children.get(c).size();
}
}
return size;
}
/**
* Is the tree empty?
*
* @return
*
* true if the tree is empty.
* false if there is still at least one value in the tree.
*/
@Override
public boolean isEmpty() {
// I am empty if I am not a leaf and I have no children
// (slightly quicker than the AbstaractCollection implementation).
return leaf == null && (children == null || children.isEmpty());
}
/**
* Returns all keys as a Set.
*
* @return
*
* A HashSet of all keys.
*
* Note: Although it returns Set<S> it is actually a Set<String> that has been
* home-grown because the original keys are not stored in the structure
* anywhere.
*/
@Override
public Set<String> keySet() {
// Roll them a temporary list and give them a Set from it.
return new HashSet<String>(keyList());
}
/**
* List all my keys.
*
* @return
*
* An ArrayList of all keys in the tree.
*
* Note: Although it returns List<S> it is actually a List<String> that has been
* home-grown because the original keys are not stored in the structure
* anywhere.
*
*/
protected List<String> keyList() {
List<String> contents = new ArrayList<String>();
if (leaf != null) {
// If I am a leaf, a null string is in the set.
contents.add((String) "");
}
// Add all sub-tries.
if (children != null) {
for (Character c : children.keySet()) {
TrieMap<V> child = children.get(c);
List<String> childContents = child.keyList();
for (String subString : childContents) {
// All possible substrings can be prepended with this character.
contents.add((String) (c + subString.toString()));
}
}
}
return contents;
}
/**
* Does the map contain the specified key.
*
* @param key
*
* The key to look for.
*
* @return
*
* true if the key is in the Map.
* false if not.
*/
public boolean containsKey(String key) {
TrieMap<V> it = find(key, false);
if (it != null) {
return it.leaf != null;
}
return false;
}
/**
* Represent me as a list.
*
* @return
*
* A String representation of the tree.
*/
@Override
public String toString() {
List<String> list = keyList();
//Collections.sort((List<String>)list);
StringBuilder sb = new StringBuilder();
Separator comma = new Separator(",");
sb.append("{");
for (String s : list) {
sb.append(comma.sep()).append(s).append("=").append(get(s));
}
sb.append("}");
return sb.toString();
}
/**
* Clear down completely.
*/
@Override
public void clear() {
children = null;
leaf = null;
}
/**
* Return a list of key/value pairs.
*
* @return
*
* The entry set.
*/
public Set<Map.Entry<String, V>> entrySet() {
Set<Map.Entry<String, V>> entries = new HashSet<Map.Entry<String, V>>();
List<String> keys = keyList();
for (String key : keys) {
entries.add(new Entry<String,V>(key, get(key)));
}
return entries;
}
/**
* An entry.
*
* @param <S>
*
* The type of the key.
*
* @param <V>
*
* The type of the value.
*/
private static class Entry<S, V> implements Map.Entry<S, V> {
protected S key;
protected V value;
public Entry(S key, V value) {
this.key = key;
this.value = value;
}
public S getKey() {
return key;
}
public V getValue() {
return value;
}
public V setValue(V newValue) {
V oldValue = value;
value = newValue;
return oldValue;
}
@Override
public boolean equals(Object o) {
if (!(o instanceof TrieMap.Entry)) {
return false;
}
Entry e = (Entry) o;
return (key == null ? e.getKey() == null : key.equals(e.getKey()))
&& (value == null ? e.getValue() == null : value.equals(e.getValue()));
}
@Override
public int hashCode() {
int keyHash = (key == null ? 0 : key.hashCode());
int valueHash = (value == null ? 0 : value.hashCode());
return keyHash^valueHash;
}
@Override
public String toString() {
return key + "=" + value;
}
}
}
快速問題：500 * 4 * 13,000,000是26,000,000,000字節或+/- 24GB - 您是否考慮將這些數據存儲在堆外？ – 2012-02-16 21:34:54
嗨500是最糟糕的情況下估計大多數字符串將只有1或2個值。現在我用-Xmx12g運行程序，但是我在另一個Map中存儲了額外的值。我悲傷的是，Map佔用了大約3g的內存和大約644mb的磁盤空間。 – samy 2012-02-16 21:43:37
Sry我沒有得到off-Heap存儲，我只是用它來搜索，這聽起來很有趣。 – samy 2012-02-16 21:55:02