Trie树判重问题

给你100000个长度不超过10的单词。对于每一个单词，我们要判断他出没出现过，如果出现了，第一次出现第几个位置。
这题当然可以用hash来，但是我要介绍的是trie树。在某些方面它的用途更大。比如说对于某一个单词，我要询问它的前缀是否出现过。这样hash就不好搞了，而用trie还是很简单。

   现在回到例子中，如果我们用最傻的方法，对于每一个单词，我们都要去查找它前面的单词中是否有它。那么这个算法的复杂度就是O(n^2)。显然对于100000的范围难以接受。现在我们换个思路想。假设我要查询的单词是abcd，那么在他前面的单词中，以b，c，d，f之类开头的我显然不必考虑。而只要找以a开头的中是否存在abcd就可以了。同样的，在以a开头中的单词中，我们只要考虑以b作为第二个字母的……这样一个树的模型就渐渐清晰了……

假设有b，abc，abd，bcd，abcd，efg，hii这6个单词，我们构建的树就是这样的。

对于每一个节点，从根遍历到他的过程就是一个单词，如果这个节点被标记为红色，就表示这个单词存在，否则不存在。 
   那么，对于一个单词，我只要顺着他从根走到对应的节点，再看这个节点是否被标记为红色就可以知道它是否出现过了。把这个节点标记为红色，就相当于插入了这个单词。

   我们可以看到，trie树每一层的节点数是26^i级别的。所以为了节省空间。我们用动态链表，或者用数组来模拟动态。空间的花费，不会超过单词数×单词长度。(转自一大牛)

Trie树的java代码 实现如下：

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


/** *//**
 * A word trie which can only deal with 26 alphabeta letters.
 * @author Leeclipse
 * @since 2007-11-21
 */

public class Trie{
 
   private Vertex root;//一个Trie树有一个根节点

    //内部类
    protected class Vertex{//节点类
        protected int words;
        protected int prefixes;
        protected Vertex[] edges;//每个节点包含26个子节点(类型为自身)
        Vertex() {
            words = 0;
            prefixes = 0;
            edges = new Vertex[26];
            for (int i = 0; i < edges.length; i++) {
                edges[i] = null;
            }
        }
    }

  
    public Trie () {
        root = new Vertex();
    }

   
    /** *//**
     * List all words in the Trie.
     * 
     * @return
     */

    public List< String> listAllWords() {
       
        List< String> words = new ArrayList< String>();
        Vertex[] edges = root.edges;
       
        for (int i = 0; i < edges.length; i++) {
            if (edges[i] != null) {
                     String word = "" + (char)('a' + i);
                depthFirstSearchWords(words, edges[i], word);
            }
        }        
        return words;
    }

     /** *//**
     * Depth First Search words in the Trie and add them to the List.
     * 
     * @param words
     * @param vertex
     * @param wordSegment
     */

    private void depthFirstSearchWords(List words, Vertex vertex, String wordSegment) {
        Vertex[] edges = vertex.edges;
        boolean hasChildren = false;
        for (int i = 0; i < edges.length; i++) {
            if (edges[i] != null) {
                hasChildren = true;
                String newWord = wordSegment + (char)('a' + i);                
                depthFirstSearchWords(words, edges[i], newWord);
            }            
        }
        if (!hasChildren) {
            words.add(wordSegment);
        }
    }

    public int countPrefixes(String prefix) {
        return countPrefixes(root, prefix);
    }

    private int countPrefixes(Vertex vertex, String prefixSegment) {
        if (prefixSegment.length() == 0) { //reach the last character of the word
            return vertex.prefixes;
        }

        char c = prefixSegment.charAt(0);
        int index = c - 'a';
        if (vertex.edges[index] == null) { // the word does NOT exist
            return 0;
        } else {

            return countPrefixes(vertex.edges[index], prefixSegment.substring(1));

        }        

    }

    public int countWords(String word) {
        return countWords(root, word);
    }    

    private int countWords(Vertex vertex, String wordSegment) {
        if (wordSegment.length() == 0) { //reach the last character of the word
            return vertex.words;
        }

        char c = wordSegment.charAt(0);
        int index = c - 'a';
        if (vertex.edges[index] == null) { // the word does NOT exist
            return 0;
        } else {
            return countWords(vertex.edges[index], wordSegment.substring(1));

        }        

    }

    
    /** *//**
     * Add a word to the Trie.
     * 
     * @param word The word to be added.
     */

    public void addWord(String word) {
        addWord(root, word);
    }

    
    /** *//**
     * Add the word from the specified vertex.
     * @param vertex The specified vertex.
     * @param word The word to be added.
     */

    private void addWord(Vertex vertex, String word) {
       if (word.length() == 0) { //if all characters of the word has been added
            vertex.words ++;
        } else {
            vertex.prefixes ++;
            char c = word.charAt(0);
            c = Character.toLowerCase(c);
            int index = c - 'a';
            if (vertex.edges[index] == null) { //if the edge does NOT exist
                vertex.edges[index] = new Vertex();
            }

            addWord(vertex.edges[index], word.substring(1)); //go the the next character
        }
    }

    public static void main(String args[])  //Just used for test
    {
    Trie trie = new Trie();
    trie.addWord("China");
    trie.addWord("China");
    trie.addWord("China");

    trie.addWord("crawl");
    trie.addWord("crime");
    trie.addWord("ban");
    trie.addWord("China");

    trie.addWord("english");
    trie.addWord("establish");
    trie.addWord("eat");
    System.out.println(trie.root.prefixes);
     System.out.println(trie.root.words);


   
     List< String> list = trie.listAllWords();
     Iterator listiterator = list.listIterator();
     
     while(listiterator.hasNext())
     {
          String s = (String)listiterator.next();
          System.out.println(s);
     }

   
     int count = trie.countPrefixes("ch");
     int count1=trie.countWords("china");
     System.out.println("the count of c prefixes:"+count);
     System.out.println("the count of china countWords:"+count1);

 
    }
}
运行:
C:\test>java   Trie
10
0
ban
china
crawl
crime
eat
english
establish
the count of c prefixes:4
the count of china countWords:4

 

文章来自:http://oivoiv.blog.163.com/