211. Design Add and Search Words Data Structure

time: O(n), n is word len

space: O(n)

class WordDictionary {
    
    class TrieNode {
        Map<Character, TrieNode> sons;
        boolean isWord;
        String word;
        TrieNode() {
            this.sons = new HashMap<>();
            this.isWord = false;
            this.word = null;
        }
    }
    
    private TrieNode root;
    public WordDictionary() {
        root = new TrieNode();
    }
    
    public void addWord(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            node.sons.computeIfAbsent(c, v -> new TrieNode());
            node = node.sons.get(c);
        }
        node.isWord = true;
        node.word = word;
    }
    
    public boolean search(String word) {
        return find(word, root, 0);
    }
    
    private boolean find(String word, TrieNode node, int index) {
        if (index == word.length()) {
            return node.isWord;
        }
        
        if (word.charAt(index) == '.') {
            for (TrieNode temp : node.sons.values()) {
                if (temp != null && find(word, temp, index+1)) { // if '.', use all sons to do DFS on word's index+1
                    return true;
                }
            }
        } else {
            // only check node's sons has this letter, if has, find on index+1
            TrieNode temp = node.sons.get(word.charAt(index)); 
            return temp != null && find(word, temp, index+1);
        }
        return false;
    }
}

/**

跟 208 不同的是, 利用 dfs, 移動 index 來找是否有這個 word
208 search 是用 iteration 的方式, 一個 char 一個 char, node = next 往下找
這裡是透過 dfs 一個一個往下找 char


 */