Question

Formatted question description: https://leetcode.ca/all/211.html

 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.

 Example:

 addWord("bad")
 addWord("dad")
 addWord("mad")
 search("pad") -> false
 search("bad") -> true
 search(".ad") -> true
 search("b..") -> true

 Constraints:
     1 <= word.length <= 500
     word in addWord consists lower-case English letters.
     word in search consist of  '.' or lower-case English letters.
     At most 50000 calls will be made to addWord and search.

Algorithm

With the structure of the Trie, the only difference is that the search function needs to be rewritten, compared with Trie problem.

Because the . in this question can replace any character, once there is a ., all subtrees need to be searched. As long as one returns true, the entire search function returns true.

Typical DFS problem.

Code

Java

public class Add_and_Search_Word_Data_structure_design {

    class WordDictionary {

        Trie trie;

        /** Initialize your data structure here. */
        public WordDictionary() {
            trie = new Trie();
        }

        /** Adds a word into the data structure. */
        public void addWord(String word) {
            trie.insert(word);
        }

        /** 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 trie.search(word);
        }
    }

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

    class TrieNode {

        // R children to node children
        public TrieNode[] children;

        private final int R = 26;

        private boolean isEnd;

        public TrieNode() {
            children = new TrieNode[R];
        }

        public boolean containsKey(char ch) {
            return children[ch -'a'] != null;
        }
        public TrieNode get(char ch) {
            return children[ch -'a'];
        }
        public void put(char ch, TrieNode node) {
            children[ch -'a'] = node;
        }
        public void setEnd() {
            isEnd = true;
        }
        public boolean isEnd() {
            return isEnd;
        }
    }

    class Trie {
        private TrieNode root;

        public Trie() {
            root = new TrieNode();
        }

        // Inserts a word into the trie.
        public void insert(String word) {
            TrieNode node = root;

            // @note:@memorize: iteration is better than recursion during trie building
            for (int i = 0; i < word.length(); i++) {
                char currentChar = word.charAt(i);
                if (!node.containsKey(currentChar)) {
                    node.put(currentChar, new TrieNode());
                }
                node = node.get(currentChar);
            }
            node.setEnd();
        }

        // search a prefix or whole key in trie and
        // returns the node where search ends
        public boolean search(String word) {

            return dfs(word, root);

        }

        private boolean dfs(String word, TrieNode node) {

            if (node == null) {
                return false;
            }

            if (word.length() == 0) {
                return node.isEnd();
            }

            char curLetter = word.charAt(0);
            if (curLetter == '.') {
                for (int i = 0; i < 26; i++) {
                    char c = (char)('a' + i);
                    if (dfs(word.substring(1), node.get(c))) {
                        return true;
                    }
                }
            } else if (node.containsKey(curLetter)){
                return dfs(word.substring(1), node.get(curLetter));
            }

            return false;
        }

    }


    public class WordDictionary_bucket {

        // bucket: word length -> list of words
        Map<Integer, List<String>> map = new HashMap<Integer, List<String>>();

        // Adds a word into the data structure.
        public void addWord(String word) {
            int index = word.length();
            if(!map.containsKey(index)){
                List<String> list = new ArrayList<String>();
                list.add(word);
                map.put(index, list);
            }else{
                map.get(index).add(word);
            }

        }

        // 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) {
            int index = word.length();
            if(!map.containsKey(index)){
                return false;
            }
            List<String> list = map.get(index);
            if(isWords(word)){
                return list.contains(word);
            }
            for(String s : list){
                if(isSame(s, word)){
                    return true;
                }
            }
            return false;
        }

        boolean isWords(String s){
            for(int i = 0; i < s.length(); i++){

                // @note: api isLetter()
                if(!Character.isLetter(s.charAt(i))){
                    return false;
                }
            }
            return true;
        }

        boolean isSame(String a, String search){
            if(a.length() != search.length()){
                return false;
            }
            for(int i = 0; i < a.length(); i++){
                if(search.charAt(i) != '.' && search.charAt(i) != a.charAt(i)){
                    return false;
                }
            }
            return true;
        }
    }

}

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