211. Add and Search Word

Problem：

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

6/5/2018 update:
这次是自己想出来的如何处理‘.’。有.时候，需要遍历当前trienode所有的children。所以用递归比较好实现。
TrieNode的children用hashmap更方便。而且居然自己想出来如何处理，for循环里面boolean结果的方法，感觉快出山了。

class WordDictionary {
    class TrieNode {
        Map<Character, TrieNode> children;
        boolean isWord;
        public TrieNode() {
            children = new HashMap<>();
        }
    }
    TrieNode root;
    /** Initialize your data structure here. */
    public WordDictionary() {
        root = new TrieNode();
    }
    
    /** Adds a word into the data structure. */
    public void addWord(String word) {
        TrieNode cur = root;
        for (char c: word.toCharArray()) {
            if (!cur.children.containsKey(c)) {
                cur.children.put(c, new TrieNode()); 
            } 
            cur = cur.children.get(c);
        }
        cur.isWord = true;
    }
    
    /** 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 helper(word, 0, root);
    }
    
    private boolean helper(String word, int index, TrieNode cur) {
        if (index == word.length()) {
            return cur.isWord;
        }
        char c = word.charAt(index);
        if (c != '.') {
            return cur.children.containsKey(c) && helper(word, index + 1, cur.children.get(c));
        } else {
            for (TrieNode node: cur.children.values()) {
                if (helper(word, index + 1, node)) {
                    return true;
                }
            }
           return false;
        }
    }
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */

--------------------------------------------------------------------------------------------------------------------------
3/6/2018 update
Analysis:
This problem is very important as it applies DFS to new field. Not the typical get a set of results DFS. But it still needs to follow the DFS template.

If DFS result is boolean, and there is a for loop to go to child nodes. We can follow the below template

            for (TrieNode n: node.children) {
                if( n != null && helper(word, n, index + 1)) return true;
            }
            return false;

Solution:

class WordDictionary {
    class TrieNode {
        char val;
        TrieNode[] children;
        boolean isWord;
        public TrieNode() {}
        public TrieNode(char val) {
            this.val = val;
            children = new TrieNode[26];
        }
    }
    TrieNode root;
    /** Initialize your data structure here. */
    public WordDictionary() {
         root = new TrieNode(' ');
    }
    
    /** Adds a word into the data structure. */
    public void addWord(String word) {
        TrieNode node = root;
        for (char c: word.toCharArray()) {
            if (node.children[c - 'a'] == null) {
                node.children[c - 'a'] = new TrieNode(c);
            }
            node = node.children[c - 'a'];
        }
        node.isWord = true;
    }
    
    /** 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 helper(word, root, 0);
    }
    
    private boolean helper(String word, TrieNode node, int index) {
        if (index == word.length()) return node.isWord;
        char c = word.charAt(index);
        if (c == '.') {
            for (TrieNode n: node.children) {
                if( n != null && helper(word, n, index + 1)) return true;
            }
            return false;
        } else {
            return node.children[c - 'a'] != null && helper(word, node.children[c - 'a'], index + 1);
        }
    }
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */