LeetCode 211. Add and Search Word

原题链接在这里：https://leetcode.com/problems/add-and-search-word-data-structure-design/

题目：

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.

题解：

查找词用到Trie树.

addword和Implement Trie (Prefix Tree)相同. search时利用dfs, 终止条件有两种.

一种是string 还没扫完, 已经走到了null节点, return false.

另一种扫完了string, 要看是否到了Trie树的叶子节点. 

dfs过程分两种情况，一种情况是当前char 是'.', 此时应该跳过这个char, 同时对剩余string和tn.nexts 的所有储存的节点继续迭代，若有一个正好走到叶子节点, 就返回ture; 若没有一个正好走到叶子节点，就return false.

另一种情况是当前char不是'.', 就继续用root.nexts[c-'a'] 和 word, i+1 index做dfs.

Time Complexity: addWord, O(n). search, O(26^n), n是string 长度. worst case每一个路径都要扫.

Space: (m*n). m is number of words added.

AC Java:

class WordDictionary {
    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 p = root;
        for(char c : word.toCharArray()){
            if(p.nexts[c - 'a'] == null){
                p.nexts[c - 'a'] = new TrieNode();
            }
            
            p = p.nexts[c - 'a'];
        }
        
        p.val = 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) {
        if(word == null || word.length() == 0){
            return false;
        }
        
        return searchWord(word, 0, root);
    }
    
    private boolean searchWord(String word, int index, TrieNode p){
        if(p == null){
            return false;
        }
        
        if(index == word.length()){
            return p.val != null;
        }
        
        if(word.charAt(index) == '.'){
            for(TrieNode next : p.nexts){
                if(searchWord(word, index + 1, next)){
                    return true;
                }
            }
            
        }else{
            return searchWord(word, index + 1, p.nexts[word.charAt(index) - 'a']);
        }
        
        return false;
    }
}

class TrieNode{
    String val;
    TrieNode [] nexts;
    
    public TrieNode(){
        nexts = new TrieNode[26];
    }
}

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