[leetcode] 211. Add and Search Word - Data structure design 解题报告

题目链接: 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.

思路: 实现一个字典树, 完成添加和查找的操作.

其原理是每个结点有26个孩子结点, 一个结点的26孩子结点初始默认为NULL. 如果我们要添加一个单词, 那么这个单词会从根节点往下一层层生长, 当到单词结尾的时候我们在最后的结点做个标记,代表从根节点到这个结点路径是一个单词.

在添加单词的时候因为公共前缀的单词会共用一些路径, 因此字典树可以极大的节约内存.

代码如下:

class WordDictionary {
public:
    typedef struct Trie
    {
        vector<Trie*> child;
        bool isWord;
        Trie(): child(vector<Trie*>(26, NULL)), isWord(false){}
    }Trie;
    
    
    WordDictionary():root(new Trie()){}
    
    // Adds a word into the data structure.
    void addWord(string word) {
        Trie* tem = root;
        for(auto ch: word)
        {
            if(tem->child[ch - 'a'] == NULL)//如果ch不在树上, 则将其添加到字典树中 
                tem->child[ch - 'a'] = new Trie();
            tem = tem->child[ch - 'a'];
        }
        tem->isWord = true;//在路径的最后标记到当前结点的路径构成一个单词
    }

    bool DFS(string str, Trie* node)
    {
        if(node == NULL) return false;
        if(str.size() ==0) return node->isWord;//如果字符串遍历完了,则看到当前结点是否可以构成一个单词
        if(str[0] != '.')//不是.则是一对一匹配
        {
            if(node->child[str[0]-'a'] == NULL) return false;
            return DFS(str.substr(1), node->child[str[0]-'a']);
        }
        for(int i = 0; i < 26; i++)//如果是.,则需要遍历所有的可能性
            if(DFS(str.substr(1), node->child[i]) == true)
                return true;
        return false;
    }

    // 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) {
        Trie* tem = root;
        return DFS(word, tem);
    }
    
private:
    Trie* root;
};

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

参考: https://leetcode.com/discuss/45435/c-using-trie-and-dfs-for-search-easy-understand-solution