TrieMap 实现

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R way Trie最大问题是占空间，R ＝ 256，插入百万数量级的key就 OutOfMemory了。在key的数目比较小的情况下，正如理论指出的一样，性能高于HashMap；但到了占内存比较多的时候由于影响系统整体运行，性能不如HashMap。

用HashMap而不是大小为256的数组作为link的时候，占内存少的多，几千万级的都不会OutOfMemory。 256-way Trie的一个Node大小至少为为256 * 8字节指针大小，也就是2K。

结论就是普通256-way Trie实用价值不大，除非确定key总数很小，空间换时间追求极致的性能。比较实用的是：

1）26-way Trie

2）用HashMap做为link容器的Trie

3）TST(Ternary Search Trie) 排序树和Trie结合起来，每个节点只有3个指针域，实现较复杂。

public class TrieMap<Value> {private static final int R = 256;private static class Node {Object value;//String prefix;int branches = 0;Node[] next = new Node[R];}private Node root = new Node();private Node put(Node x, String key, int i, Value v) {if (x == null) x = new Node();//x.prefix = key.substring(0, i);if (i == key.length()) {x.value = v;return x;}Node link = x.next[key.charAt(i)];x.next[key.charAt(i)] = put(x.next[key.charAt(i)], key, i + 1, v);if (link == null && x.next[key.charAt(i)] != null) x.branches += 1;return x;}public void put(String key, Value value) { root = put(root, key, 0, value);}private Node get(Node x, String key, int i) {if (x == null) return null;if (i == key.length()) return x;return get(x.next[key.charAt(i)], key, i + 1);}public Value get(String key) { Node x = get(root, key, 0);return x == null ? null : (Value)x.value;}private Node remove(Node x, String key, int i) {if (x == null) return null;if (i == key.length()) {x.value = null;if (x.branches == 0) return null;return x;}Node link = x.next[key.charAt(i)];x.next[key.charAt(i)] = remove(x.next[key.charAt(i)], key, i + 1);if (link != null && x.next[key.charAt(i)] == null) x.branches -= 1;if (x.value == null && x.branches == 0) return null;return x;}public void remove(String key) { remove(root, key, 0);}private void collect (Node root, String prefix, List<String> list) {if (root == null) return;if (root.value != null) list.add(prefix);for (char c = 0; c < R; ++c) {collect(root.next[c], prefix + c, list);}}public List<String> keys() {return keysWithPrefix("");}public List<String> keysWithPrefix(String prefix) {Node x = get(root, prefix, 0);List<String> list = new ArrayList<String>();collect(x, prefix, list);return list;}int search (Node x, String query, int i, int length) {if (x == null) return length;if (x.value != null) length = i;if (i == query.length()) return length;return search(x.next[query.charAt(i)], query, i + 1, length);}public String longestPrefixOf(String query) {int len = search(root, query, 0, 0);return query.substring(0, len);}private String maxKey(Node x, String prefix) {if (x == null) return null;for (char c = R - 1; c >= 0; --c) {String s = maxKey(x.next[c], prefix + c);if (s != null) return s;}if (x.value != null) return prefix;return null;}}

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