LeetCode: 211.Add and Search Word - Data structure design

题目描述

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.

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.

解题思路

参考 LeetCode: 208. Implement Trie (Prefix Tree) 题解。当遇到 . 字符时，遍历所有的可能性。

AC 代码

class WordDictionary {
private:
    const static int CHILD_NODE_NUM = 27;
    struct TrieNode
    {
        TrieNode* ch[CHILD_NODE_NUM];
        size_t count;
        
        TrieNode():count(0)
        {
            for(size_t i = 0; i < CHILD_NODE_NUM; ++i)
            {
                ch[i] = nullptr;
            }
        }
    };
public:
    /** Initialize your data structure here. */
    WordDictionary() {
        root = new TrieNode();
    }
    
    /** Adds a word into the data structure. */
    void addWord(string word) {
        insert(root, word, 0);
    }
    
    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    bool search(string word) {
        return search(root, word, 0);
    }
private:
    void insert(TrieNode* r, string word, int idx)
    {
        int chIdx = CHILD_NODE_NUM-1;
      
        if(word.size() < idx)
        {
            return;
        }
        else if(word.size() != idx)
        {
            chIdx = word[idx]-'a';
        }
        
        if(r->ch[chIdx] == nullptr)
        {
            r->ch[chIdx] = new TrieNode();
        }
        
        ++(r->ch[chIdx]->count);
        
        insert(r->ch[chIdx], word, idx+1);
    }
    
    bool search(TrieNode* r, string word, int idx)
    {
        int chIdx = CHILD_NODE_NUM-1;
      
        if(word.size() < idx)
        {
            return r->count;
        }
        else if(word[idx] == '.' && word.size() != idx)
        {
            bool bRet = false;
            for(int i = 0; !bRet && i < CHILD_NODE_NUM; ++i)
            {
                if(r->ch[i]) bRet = search(r->ch[i], word, idx+1);
            }
            
            return bRet;
        }
        else if(word.size() != idx)
        {
            chIdx = word[idx]-'a';
        }
        
        if(r->ch[chIdx] == nullptr)
        {
            return false;
        }
        
        return search(r->ch[chIdx], word, idx+1);
    }
private:
    TrieNode* root;
};

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