Leetcode/F家 -- 211. Add and Search Word - Data structure design (Trie+ Backtracking)

211. Add and Search Word (Trie+Backtracking)

• Difficulty: Medium

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.

For 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.

相关题： http://rainykat.blogspot.com/2017/01/leetcode-208-implement-trie-prefix-tree.html

思路：按照Trie写data structure, when meet '.', we need do back tracking. 
          Base case is index == chs.length since root is NULL
Complexity: worst case of all '...' it will check all nodes of trie,  O(26^n) n is key length
关键字： dfs， Trie 

public class WordDictionary {
    //data structure
    public class TrieNode{
        public TrieNode[] links = new TrieNode[26];
        public boolean isEnd;
    }
    private TrieNode root = new TrieNode();
    
    // Adds a word into the data structure.
    public void addWord(String word) {
        TrieNode node = root;
        for(int i=0;i<word.length();i++){
            char cur = word.charAt(i);
            if(node.links[cur-'a']==null) node.links[cur-'a'] = new TrieNode();
            node = node.links[cur-'a'];
        }
        node.isEnd = 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) {
        //need backtracking
        return match(word.toCharArray(),0,root);
    }
    
    public boolean match(char[] chs,int index,TrieNode node){
        if (index == chs.length) return node.isEnd;//because root is NULL
    
        if (chs[index] != '.') {
           return node.links[chs[index] - 'a'] != null && match(chs, index + 1, node.links[chs[index] - 'a']);
        } else {//backtracking: set of options which is links[i],try each recursively
             for (int i = 0; i < node.links.length; i++) {//recur any links of '.'
                if (node.links[i] != null && match(chs, index + 1, node.links[i])) {
                    return true;//if there is match, we just return true
                }//if not match, we just skip to next one, until no char left and we ensure no match
            }
        }
        return false;
    }
}

// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");