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.

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.

click to show hint.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

解题思路：

这题实际上是 Implement Trie (Prefix Tree) 的follow-up。有了上一题的设计，这题唯一需要解决的，就是search这个方法，当遇到'.'的时候，如何办。

本题的addword()方法中，是没有'.'的，所以Trie树内肯定没有'.'。那么当遇到'.'的时候，只要递归search下一层次的所有子节点，有一个路径里找到，就返回true，否则立刻返回false。

public class WordDictionary {
    class TrieNode {
        // Initialize your data structure here.
        boolean isWord;
        Map<Character, TrieNode> next;
        public TrieNode() {
            next = new HashMap<Character, TrieNode>();
            isWord = false;
        }
    }
    
    private TrieNode root;

    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++) {
            if(cur.next.get(word.charAt(i)) == null) {
                TrieNode next = new TrieNode();
                cur.next.put(word.charAt(i), next);
            }
            cur = cur.next.get(word.charAt(i));
        }
        cur.isWord = 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, word);
    }
    
    public boolean searchHelper(TrieNode cur, String word) {
        if(cur == null) {
            return false;
        }
        for(int i = 0; i < word.length(); i++) {
            if(word.charAt(i) != '.') {
                cur = cur.next.get(word.charAt(i));
            } else {
                for(TrieNode value : cur.next.values()) {
                    if(searchHelper(value, word.substring(i + 1))) {
                        return true;
                    }
                }
                return false;
            }
            if(cur == null) {
                return false;
            }
        }
        return cur.isWord;
    }
}

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

 2018/6/20 二刷

class TrieNode {
    HashMap<Character, TrieNode> node;
    boolean isWord;
    
    public TrieNode() {
        node = new HashMap<Character, TrieNode>();
        isWord = false;
    }
}

public class WordDictionary {

    /** 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++) {
            if (!cur.node.containsKey(word.charAt(i))) {
                TrieNode node = new TrieNode();
                cur.node.put(word.charAt(i), node);
            }
            cur = cur.node.get(word.charAt(i));
        }
        cur.isWord = 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, word);
    }
    
    public boolean searchHelper(TrieNode cur, String word) {
        for (int i = 0; i < word.length(); i++) {
            if (word.charAt(i) == '.') {
                // 这里是所有都没找到才返回false，不能直接return searchHelper(cur.node.get(key), word.substring(i + 1))
                for (Character key : cur.node.keySet()) {
                    if (searchHelper(cur.node.get(key), word.substring(i + 1))) {
                        return true;
                    }
                }
                return false;
            }
            if (!cur.node.containsKey(word.charAt(i))) {
                return false;
            }
            cur = cur.node.get(word.charAt(i));
        }
        return cur.isWord;
    }
    
    private TrieNode root;
}

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