211. 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

一刷
用trie字典树和DFS来解决

public class WordDictionary {
    private TrieNode root;
    
    public class TrieNode {
        public TrieNode[] children = new TrieNode[26];
        public boolean endOfWord = false;
    }

    /** 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();
            }
            node = node.children[c - 'a'];
        }
        node.endOfWord = 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 match(word.toCharArray(), 0, root);
    }
    
    private boolean match(char[] chs, int k, TrieNode node) {
        if (k == chs.length) return node.endOfWord;   
        if (chs[k] != '.') {
            return node.children[chs[k] - 'a'] != null && match(chs, k + 1, node.children[chs[k] - 'a']);
        } else {
            for (int i = 0; i < node.children.length; i++) {
                if (node.children[i] != null) {
                    if (match(chs, k + 1, node.children[i])) {
                        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);
 */

二刷
字典树

public class WordDictionary {

    private class TrieNode{
        TrieNode[] child;
        boolean endOfWord;
        
        TrieNode(){
            child = new TrieNode[26];
        }
        
    }
    
    private 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(int i=0; i<word.length(); i++){
            char ch = word.charAt(i);
            if(cur.child[ch - 'a'] == null){
                cur.child[ch - 'a'] = new TrieNode();
            }
            cur = cur.child[ch - 'a'];
        }
        cur.endOfWord = 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 search(word.toCharArray(), root, 0);
    }
    
    public boolean search(char[] word, TrieNode cur, int k){
        if(k == word.length) return cur.endOfWord;
        if(word[k]!='.'){
            if(cur.child[word[k] - 'a'] == null) return false;
            else return search(word, cur.child[word[k] - 'a'], k+1);
        }else{
            for(int i=0; i<26; i++){
                if(cur.child[i]!=null){
                    if(search(word, cur.child[i], k+1)) 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);
 */