字典树

Design a data structure that supports the following two operations:

void addWord(word)bool search(word)

search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

For example:

addWord("bad")addWord("dad")addWord("mad")search("pad") -> falsesearch("bad") -> truesearch(".ad") -> truesearch("b..") -> true

Note:

You may assume that all words are consist of lowercase letters a-z.

public class WordDictionary {class TrieNode {    final static int maxnode = 40000;    final static int sigma_size = 26;    // Initialize your data structure here.    int ch[][] = new int[maxnode][sigma_size];    int val[] = new int[maxnode];    int sz = 1; // 结点总数    int idx(char c) {        return c - 'a';    } // 字符c的编号    // 插入字符串s，附加信息为v。注意v必须非0，因为0代表“本结点不是单词结点”    void insert(String s, int v) {        int u = 0, n = s.length();        for (int i = 0; i < n; i++) {            int c = idx(s.charAt(i));            if (ch[u][c] == 0) { // 结点不存在                //memset(ch[sz], 0, sizeof(ch[sz]));                val[sz] = 0;  // 中间结点的附加信息为0                ch[u][c] = sz++; // 新建结点            }            u = ch[u][c]; // 往下走        }        val[u] = v;    }    int find(String s)//查找s，返回权值    {        int u = 0;        for (int i = 0; i < s.length(); i++) {            int c = idx(s.charAt(i));            if (ch[u][c] == 0) return -1;            u = ch[u][c];        }        return val[u];    }    int findRescure(String s, int u, int start) {        if (start == s.length()) return val[u];        int c = idx(s.charAt(start));        if (ch[u][c] == 0) return -1;        return findRescure(s, ch[u][c], start + 1);    }   int findWithDot(String s, int u, int start) {        if (start == s.length()) {            return val[u];        }        if (s.charAt(start) == '.') {            for (int j = 0; j < 26; j++) {                if (ch[u][j] == 0) continue;                int res = findWithDot(s, ch[u][j], start + 1);                if (res != -1 && res != 0)                    return res;            }        } else {            int c = idx(s.charAt(start));            if (ch[u][c] == 0) {                return -1;            }            return findWithDot(s, ch[u][c], start + 1);        }        return -1;    }    public TrieNode() {    }}public class Trie {    private TrieNode root;    public Trie() {        root = new TrieNode();    }    // Inserts a word into the trie.    public void insert(String word) {        root.insert(word, 1);    }    // Returns if the word is in the trie.    public boolean search(String word) {        int res = root.findWithDot(word,0,0);        return res != -1 && res != 0;    }    // Returns if there is any word in the trie    // that starts with the given prefix.    public boolean startsWith(String prefix) {        return root.find(prefix) != -1;    }}        Trie trie = new Trie();    // Adds a word into the data structure.    public void addWord(String word) {        trie.insert(word);    }    // Returns if the word is in the data structure. A word could    // contain the dot character '.' to represent any one letter.    public boolean search(String word) {        return trie.search(word);    }}// Your WordDictionary object will be instantiated and called as such:// WordDictionary wordDictionary = new WordDictionary();// wordDictionary.addWord("word");// wordDictionary.search("pattern");