Java:使前缀树记住最后一个不为空的值

Question

我有这样的前缀树(Trie)代码:

public class Trie<V> {

Entry<V> entry;
char key;
Map<Character, Trie<V>> childrens;

public Trie() {
    this.childrens = new HashMap<Character, Trie<V>>(10);
    entry = new Entry<V>();
}

/** non-public, used by _put() */
Trie(char key) {
    this.childrens = new HashMap<Character, Trie<V>>(10);
    this.key = key;
    entry = new Entry<V>();
}

public void put(String key, V value) {
    _put(new StringBuffer(key), new StringBuffer(""), value);
}

void _put(StringBuffer remainder, StringBuffer prefix, V value) {
    if (remainder.length() > 0) {
        char keyElement = remainder.charAt(0);
        Trie<V> t = null;
        try {
            t = childrens.get(keyElement);
        } catch (IndexOutOfBoundsException e) {
        }
        if (t == null) {
            t = new Trie<V>(keyElement);
            childrens.put(keyElement, t);
        }
        prefix.append(remainder.charAt(0));
        t._put(remainder.deleteCharAt(0), prefix, value);
    } else {
        this.entry.value = value;
        this.entry.prefix = prefix.toString();
    }

}

/**
 * Retrieves element from prefix table matching as a prefix to provided key.
 * E.g. is key is "abcde" and prefix table has node "ab" then this call will
 * return "ab"
 * 
 * @param key
 *            a string which starts with prefix to be searched in the table
 *            (e.g. phone number)
 * @return an Object assosiated with matching prefix (i.e if key is a phone
 *         number it may return a corresponding country name)
 */
public V get(String key) {
    return _get(new StringBuffer(key), 0);
}

/**
 * Returns true if key has matching prefix in the table
 */
public boolean hasPrefix(String key) {
    return ((this.get(key) != null) ? true : false);
}

V _get(StringBuffer key, int level) {
    if (key.length() > 0) {
        Trie<V> t = childrens.get(key.charAt(0));
        if (t != null) {
            return t._get(key.deleteCharAt(0), ++level);
        } else {
            return (level > 0) ? entry.value : null;
        }

    } else {
        return entry.value;
    }
}

@Override
public String toString() {
    return "Trie [entry=" + entry + ", key=" + key + ", childrens="
            + childrens + "]";
}

static public class Entry<V> {
    String prefix;
    V value;

    public Entry() {
    }

    public Entry(String p, V v) {
        prefix = p;
        value = v;
    }

    public String prefix() {
        return prefix;
    }

    public V value() {
        return value;
    }

    @Override
    public String toString() {
        return "Entry [prefix=" + prefix + ", value=" + value + "]";
    }

}
}

插入是这样的:

private static Trie<String> trie = new Trie<>();
trie.put("7", "Some country");
trie.put("77", "Some other country");
trie.put("745", "Entirely different place");

搜索将如下所示:

String result = trie.get("746878788");
System.out.println(result);

此搜索结果将为 null，因为没有 74 的值。

我的问题是:如何修改 Trie 类中的 _get 方法，以便它记住最后一个不为空的值。因此，当它最终为 74 时，它会记住 7 有某个值“某个国家/地区”，因此它将返回该值而不是 null。

有什么想法可以解决这个问题吗？

感谢任何帮助!

Answer 1

首先，我修改了 Trie 的 toString() 方法以获得一些更好的调试信息。实现您所要求的唯一重要的行是 _get 方法中的这些:

V result = t._get(key.deleteCharAt(0), ++level);
return result == null ? entry.value : result;

如果子条目的值为 null，则字典树现在更喜欢当前条目的值。

修改后的整个代码:

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;

public class Trie<V> {

    Entry<V> entry;
    char key;
    Map<Character, Trie<V>> children;

    public Trie() {
        this.children = new HashMap<Character, Trie<V>>(10);
        entry = new Entry<V>();
    }

    /** non-public, used by _put() */
    Trie(char key) {
        this.children = new HashMap<Character, Trie<V>>(10);
        this.key = key;
        entry = new Entry<V>();
    }

    public void put(String key, V value) {
        _put(new StringBuffer(key), new StringBuffer(""), value);
    }

    void _put(StringBuffer remainder, StringBuffer prefix, V value) {
        if (remainder.length() > 0) {
            char keyElement = remainder.charAt(0);
            Trie<V> t = null;
            try {
                t = children.get(keyElement);
            } catch (IndexOutOfBoundsException e) {
            }
            if (t == null) {
                t = new Trie<V>(keyElement);
                children.put(keyElement, t);
            }
            prefix.append(remainder.charAt(0));
            t._put(remainder.deleteCharAt(0), prefix, value);
        } else {
            this.entry.value = value;
            this.entry.prefix = prefix.toString();
        }

    }

    /**
     * Retrieves element from prefix table matching as a prefix to provided key.
     * E.g. is key is "abcde" and prefix table has node "ab" then this call will
     * return "ab"
     * 
     * @param key
     *            a string which starts with prefix to be searched in the table
     *            (e.g. phone number)
     * @return an Object assosiated with matching prefix (i.e if key is a phone
     *         number it may return a corresponding country name)
     */
    public V get(String key) {
        return _get(new StringBuffer(key), 0);
    }

    /**
     * Returns true if key has matching prefix in the table
     */
    public boolean hasPrefix(String key) {
        return ((this.get(key) != null) ? true : false);
    }

    V _get(StringBuffer key, int level) {
        if (key.length() > 0) {
            Trie<V> t = children.get(key.charAt(0));
            if (t != null) {
                 V result = t._get(key.deleteCharAt(0), ++level);
                 return result == null ? entry.value : result;

            } else {
                return (level > 0) ? entry.value : null;
            }

        } else {
            return entry.value;
        }
    }

    @Override
    public String toString() {

        Iterator<Character> it = children.keySet().iterator();
        StringBuffer childs = new StringBuffer();

        while (it.hasNext()) {
            Character key = it.next();
            childs.append(String.format("\n%s\n",
                    // adding a tab to the beginning of every line to create a visual tree
                    String.format("%s: %s", key, children.get(key)).toString().replaceAll("(?m)(^)", "\t")));
        }

        return String.format("Trie [entry=%s, children=%s]", entry, childs);
    }

    static public class Entry<V> {
        String prefix;
        V value;

        public Entry() {
        }

        public Entry(String p, V v) {
            prefix = p;
            value = v;
        }

        public String prefix() {
            return prefix;
        }

        public V value() {
            return value;
        }

        @Override
        public String toString() {
            return "Entry [prefix=" + prefix + ", value=" + value + "]";
        }

    }
}