211. 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.

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/

题解：

设计一个Data Structure来search和add单词。这道题我们又可以用一个R-Way Trie来完成。 像JQuery里面的Auto-complete功能其实就可以用R-Way Trie based method来设计和编程。注意当字符为"."的时候我们要loop当前节点的全部26个子节点，这里要用一个DFS。

Time Complexity - O(n)，  Space Complextiy - O(26ⁿ)。

public class WordDictionary {
    private TrieNode root = new TrieNode();
    
    private class TrieNode {
        private final int R = 26;           // radix = 26
        public TrieNode[] next;
        public boolean isWord;
        
        public TrieNode() {
            next = new TrieNode[R];
        }
    }

    // Adds a word into the data structure.
    public void addWord(String word) {
        if(word == null || word.length() == 0)
            return;
        TrieNode 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 TrieNode();
            node = node.next[c - 'a'];
            d++;
        }
        
        node.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) {
        if(word == null || word.length() == 0)
            return false;
        TrieNode node = root;
        int d = 0;
        
        return search(node, word, 0);
    }
    
    private boolean search(TrieNode node, String word, int d) {
        if(node == null)
            return false;
        if(d == word.length())
            return node.isWord;
        char c = word.charAt(d);
        if(c == '.') {
            for(TrieNode child : node.next) {
                if(child != null && search(child, word, d + 1))
                    return true;
            }
            return false;
        } else {
            return search(node.next[c - 'a'], word, d + 1);
        }
    }
}

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

 

二刷:

方法和一刷一样，主要使用Trie。addWord的时候还是使用和Trie的insert一样的的代码。 Search的时候因为有一个通配符'.'，所以我们要用dfs搜索节点的26个子节点。

假如使用Python的话可以不用Trie，直接用dict来做。

Java:

Time Complexity:  addWord - O(L) ,   search - O(26^L)，  Space Complexity - O(26^L)   这里 L是单词的平均长度。

 

public class WordDictionary {
    TrieNode root = new TrieNode();
    // Adds a word into the data structure.
    public void addWord(String word) {
        if (word == null) return;
        TrieNode node = this.root;
        int d = 0;
        while (d < word.length()) {
            int index = word.charAt(d) - 'a';
            if (node.next[index] == null) node.next[index] = new TrieNode();
            node = node.next[index];
            d++;
        }
        node.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 search(word, root, 0);
    }
    
    private boolean search(String word, TrieNode node, int depth) {
        if (node == null) return false;
        if (depth == word.length()) return node.isWord;
        char c = word.charAt(depth);
        if (c != '.') {
            return search(word, node.next[c - 'a'], depth + 1);
        } else {
            for (TrieNode nextNode : node.next) {
                if (search(word, nextNode, depth + 1)) return true;
            }
            return false;
        }
    }
    
    private class TrieNode {
        TrieNode[] next;
        int R = 26;
        boolean isWord;
        
        public TrieNode() {
            this.next = new TrieNode[R];
        }
    }
}

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

 

 

Reference:

https://leetcode.com/discuss/35878/java-hashmap-backed-trie

https://leetcode.com/discuss/35928/my-simple-and-clean-java-code

https://leetcode.com/problems/implement-trie-prefix-tree/

https://leetcode.com/discuss/69963/python-168ms-beat-100%25-solution