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 Trie100banchinacrawlcrimeeatenglishestablishthe count of c prefixes:4the count of china countWords:4?
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