211. Add and Search Word

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.

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.

链接： http://leetcode.com/problems/add-and-search-word-data-structure-design/

7/14/2017

58%，用Trie

参考508官方解答做的addWord，参考别人答案做的search，search部分主要是碰见'.'则遍历所有26个字母，是DFS，所以要把search分开，其中第二个可以用recursive写比较方便。

public class WordDictionary {
    private static final int R = 26;
    private Node root;
    
    private static class Node {
        private boolean inWord;
        private Node[] next = new Node[R];
    }
    /** Initialize your data structure here. */
    public WordDictionary() {
        root = new Node();
    }
    
    /** Adds a word into the data structure. */
    public void addWord(String word) {
        if (word == null || word.length() == 0) {
            return;
        }
        Node node = root;
        int d = 0;
        while (d < word.length()) {
            char c = word.charAt(d);
            if(node.next[c - 'a'] == null) {
                node.next[c - 'a'] = new Node();
            }
            node = node.next[c - 'a'];
            d++;
        }
        node.inWord = 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) {
        if (word == null || word.length() == 0) {
            return false;
        }
        return search(root, word, 0);
    }
    private boolean search(Node x, String key, int d) {
        if (x == null) {
            return false;
        }
        if (d == key.length()) {
            return x.inWord;
        }
        char c = key.charAt(d);
        if (c == '.') {
            for (Node node: x.next) {
                if (node != null && search(node, key, d + 1)) {
                    return true;
                }
            }
            return false;
        } else {
            if (x.next[c - 'a'] == null) {
                return false;
            }
            return search(x.next[c - 'a'], key, d + 1);
        }
    }
}

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

别人的答案

http://www.cnblogs.com/yrbbest/p/4979621.html

https://discuss.leetcode.com/topic/14216/tree-solutions-18-20-lines

更多讨论

https://discuss.leetcode.com/category/219/add-and-search-word-data-structure-design