LeetCode 211 - Add and Search Word - Data structure design

https://leetcode.com/problems/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

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/discuss/59554/My-simple-and-clean-Java-code

Using backtrack to check each character of word to search.

public class WordDictionary {
    public class TrieNode {
        public TrieNode[] children = new TrieNode[26];
        public String item = "";
    }
    
    private TrieNode root = new TrieNode();

    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.item = word;
    }

    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.item.equals("");   
        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;
    }

https://github.com/mintycc/OnlineJudge-Solutions/blob/master/Leetcode/211_Add_and_Search_Word_Data_Structure_Design.java

    class TrieNode {
        boolean finished;
        TrieNode[] children;
        TrieNode() {
            finished = false;
            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 cur = root;
        for (char ch : word.toCharArray()) {
            if (cur.children[ch - 'a'] == null)
                cur.children[ch - 'a'] = new TrieNode();
            cur = cur.children[ch - 'a'];
        }
        cur.finished = 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 searchHelper(root, 0, word);
    }
    
    private boolean searchHelper(TrieNode root, int index, String word) {
        if (index == word.length()) return root.finished;
        char ch = word.charAt(index);
        if (ch != '.') {
            if (root.children[ch - 'a'] == null) return false;
            return searchHelper(root.children[ch - 'a'], index + 1, word);
        } else {
            for (int i = 0; i < 26; i ++)
                if (root.children[i] != null)
                    if (searchHelper(root.children[i], index + 1, word))
                        return true;
            return false;
        }
    }

follow up是虽然这个已经是optimal的⽅法，有没有可能让search更更快⼀一
点，表示在insert的时候就把“.”也当做⼀一个node加进去，⽐比如如果insert
“cat”, 那么第⼀一 层就是 “c”,".",第⼆二层是“a”,"." | "a","."

讨论：这道题有个很好的Follow up，就是当搜索的单词中存在星号怎么搞，星号的定义和Wildcard Matching中一样，可以代表任意的字符串，包括空字符串，请参见评论区1楼

if (word[i] == '*') {

    if (i + 1 == word.size()) return true;

    for (auto &a : p->child) {

        if (a && (a->child[word[i + 1] - 'a'] || word[i + 1] == '.' || word[i + 1] == '*') && searchWord(word, a, i + 1)) return true;

        if (a && searchWord(word, a, i)) return true;

    }

    return false;

}