【两次过】Lintcode 473. 单词的添加与查找

设计一个包含下面两个操作的数据结构：addWord(word), search(word)

addWord(word)会在数据结构中添加一个单词。而search(word)则支持普通的单词查询或是只包含.和a-z的简易正则表达式的查询。

一个 . 可以代表一个任何的字母。

样例

样例 1:

输入:
  addWord("a")
  search(".")
输出: 
  true

样例 2:

输入: 
  addWord("bad")
  addWord("dad")
  addWord("mad")
  search("pad")  
  search("bad")  
  search(".ad")  
  search("b..") 
输出:
  false
  true
  true
  true 

注意事项

你可以认为所有的单词都只包含小写字母 a-z。

---

解题思路1：

与Lintcode 442. 实现 Trie（前缀树）类似。不同的是增加了对于‘.’正则表达式的处理，需要搜索每一个可能性，循环方法

public class WordDictionary {
    private class Node{
        public boolean isWord;  //说明以当前节点为尾的字符串能否组成一个单词
        public Map<Character, Node> next;   //next用map存储连接当前节点的下一个节点的值的集合
        
        public Node(boolean isWord){
            this.isWord = isWord;
            this.next = new HashMap<>();
        }
        
        public Node(){
            this(false);
        }
    }
    
    private Node root;
    
    public WordDictionary(){
        root = new Node();
    }
    /*
     * @param word: Adds a word into the data structure.
     * @return: nothing
     */
    public void addWord(String word) {
        // write your code here
        Node cur = root;
        
        for(int i=0; i<word.length(); i++){
            char c = word.charAt(i);
            
            if(cur.next.get(c) == null)
                cur.next.put(c, new Node());
            
            cur = cur.next.get(c);
        }
        
        cur.isWord = true;
    }

    /*
     * @param word: A word could contain the dot character '.' to represent any one letter.
     * @return: if the word is in the data structure.
     */
    public boolean search(String word) {
        // write your code here
        Node cur = root;
        
        for(int i=0; i<word.length(); i++){
            char c = word.charAt(i);
            if(c == '.'){
                //处理正则表达式，需搜索当前节点的next中的每一种可能性
                for(Character ch : cur.next.keySet()){
                    if(search(word.substring(0,i) + ch + word.substring(i+1)))
                        return true;
                }
                return false;
            }
            
            if(cur.next.get(c) == null)
                return false;
            
            cur = cur.next.get(c);
        }
        
        return cur.isWord;
    }
    
}

解题思路2：

递归方法。

public class WordDictionary {
    private class Node{
        public boolean isWord;  //说明以当前节点为尾的字符串能否组成一个单词
        public Map<Character, Node> next;   //next用map存储连接当前节点的下一个节点的值的集合
        
        public Node(boolean isWord){
            this.isWord = isWord;
            this.next = new HashMap<>();
        }
        
        public Node(){
            this(false);
        }
    }
    
    private Node root;
    
    public WordDictionary(){
        root = new Node();
    }
    /*
     * @param word: Adds a word into the data structure.
     * @return: nothing
     */
    public void addWord(String word) {
        // write your code here
        Node cur = root;
        
        for(int i=0; i<word.length(); i++){
            char c = word.charAt(i);
            
            if(cur.next.get(c) == null)
                cur.next.put(c, new Node());
            
            cur = cur.next.get(c);
        }
        
        cur.isWord = true;
    }

    /*
     * @param word: A word could contain the dot character '.' to represent any one letter.
     * @return: if the word is in the data structure.
     */
    public boolean search(String word) {
        // write your code here
        return match(root, word, 0);
    }
    
    //考察word中从index开始的字符串是否匹配以node为根的next节点
    private boolean match(Node node, String word, int index){
        if(index > word.length())
            return false;
        
        if(index == word.length())
            return node.isWord;
        
        char c = word.charAt(index);
        
        if(c != '.'){
            if(node.next.get(c) == null)
                return false;
            return match(node.next.get(c), word, index+1);
            
        }else{  //c == '.'
            
            for(Character ch : node.next.keySet()){
                if(match(node.next.get(ch), word, index+1))
                    return true;
            }
            return false;
        }
            
        
            
    }
    
}