[LeetCode][JavaScript]Add and Search Word

Add and Search Word - Data structure design

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") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

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

https://leetcode.com/problems/add-and-search-word-data-structure-design/

---

字典树，在上一题的基础上稍做修改。http://www.cnblogs.com/Liok3187/p/4626730.html

遇到'.'就要采用递归的方式，遍历这一层词典里所有的词。

/**
 * @constructor
 */
var WordDictionary = function() {
    this.root = new TrieNode('root');
};

function TrieNode(key) {
    return {
        key : key,  
        isWord : false
    };
}

/**
 * @param {string} word
 * @return {void}
 * Adds a word into the data structure.
 */
WordDictionary.prototype.addWord = function(word) {
    var tree = this.root, i, curr;
    for(i = 0; i < word.length; i++){
        curr = word[i];
        if(!tree[curr]){
            tree[curr] = new TrieNode(curr);
        }
        tree = tree[curr];
    }
    tree.isWord = true;
};

/**
 * @param {string} word
 * @return {boolean}
 * Returns if the word is in the data structure. A word could
 * contain the dot character '.' to represent any one letter.
 */
WordDictionary.prototype.search = function(word) {
    return searchWord(word, this.root);

    function searchWord(word, tree){
        if(word === "" && tree.isWord){
            return true;
        }
        if(word[0] !== '.'){
            if(!tree[word[0]]){
                return false;
            }else{
                return searchWord(word.substring(1, word.length), tree[word[0]]);
            }
        }else{
            for(var i in tree){
                if(i === 'key' || i === 'isWord'){
                    continue;
                }
                if(searchWord(word.substring(1, word.length), tree[i])){
                    return true;
                }
            }
            return false;
        }
    }
};