lintcode 473. Add and Search Word - Data structure design

典型的字典树trie题
链接
字典树结构就不再详述，这里的addword操作就如同常规的字典树增加单词的操作。

这里的查询操作有所不同，出现了'.'， 这个符号可以代表任意的26字母，那么在查询的时候就要涉及到了递归，如果是字母的话常规查询即可，但一旦遇到通配符的时候，要遍历26个子树，任意一个树满足统通配符及其后面的字符即代表查询成功。这里涉及到了或运算。

代码如下

public class WordDictionary {
    Trie t = new Trie();
    /*
     * @param word: Adds a word into the data structure.
     * @return: nothing
     */
    public void addWord(String word) {
        // write your code here
        Trie p = t;
        for(int i = 0; i < word.length(); i++){
            int idx = word.charAt(i) - 'a';
            if(p.subTries[idx] == null){
                p.subTries[idx] = new Trie();
            }
            p = p.subTries[idx];
        }
        p.end = true;
    }

    /*
     * @param word: A word could contain the dot character '.' to represent any one letter.
     * @return: if the word is in the data structure.
     */
    public boolean search(String word) {
        // write your code here
        return searchW(word, t);
        
    }
    
    private boolean searchW(String word, Trie root){
        Trie p = root;
        
        for(int i = 0; i < word.length(); i++){
            if(word.charAt(i) == '.'){
                boolean res = false;
                for(int j = 0; j < 26; j++){
                    if(p.subTries[j] != null){
                        
                        res |= searchW(word.substring(i+1), p.subTries[j]);
                    }
                }
                return res;
            }else{
                int idx = word.charAt(i) - 'a';
                if(p.subTries[idx] == null){
                    return false;
                }
                p = p.subTries[idx];
            }
        }
        return p.end == true;
    }
    
}
class Trie{
    Trie[] subTries;
    boolean end;
    public Trie(){
        end = false;
        subTries = new Trie[26];
    }
    
}