leetcode 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.

题目来源：https://leetcode.com/problems/add-and-search-word-data-structure-design/

分析

关于字典树（trie tree）的概念和分析见：/article/2077132.html

主要挑战性在于’.’的匹配。遇到一个’.’就得匹配当前next[]数组里面所有的有效分支。用个递归函数，递归匹配吧。

代码

[code]class TrieNode{
public:
    int flag;
    TrieNode* next[26];

    TrieNode(){
        flag = 0;
        for(int i = 0; i < 26; i++){
            next[i] = NULL;
    }
}
};
class WordDictionary {
private:
    TrieNode *root;
public:
    WordDictionary(){
        root = new TrieNode();
    }
    // Adds a word into the data structure.
    void addWord(string word) {
        TrieNode* p = root;
        int len = word.length();
        if(len <= 0)
            return;
        for(int i = 0; i < len; i++){
            int d = word.at(i) - 'a';
            if((p->next)[d] == NULL){
                p->next[d] = new TrieNode();
                p->next[d]->flag = 0;
            }
            p = p->next[d];
        }
        p->flag = 1;
    }

    // Returns if the word is in the data structure. A word could
    // contain the dot character '.' to represent any one letter.
    bool searchHelp(string word, struct TrieNode *root){
        struct TrieNode *p = root;
        int len = word.length();
        if(len <= 0){//为了匹配字符串中字符'.'是最后一个字符的情形。
            if(p->flag == 1)
                return true;
            else
                return false;
        }
        if(root == NULL)
            return false;
        for(int i = 0; i < len; i++){
            if(word.at(i) == '.'){
                bool result = false;
                for(int j = 0; j < 26; j++){
                    if(p->next[j] != NULL){
                        result = result || searchHelp(word.substr(i+1), p->next[j]);
                    }
                }
                return result;
            }
            else
            {
                int d = word.at(i) - 'a';
                if(p->next[d] == NULL)
                    return false;
                p = p->next[d];
            }
        }
        if(p->flag == 1)
                return true;
            else
                return false;
    }
    bool search(string word) {
        int len = word.length();
        if(len <= 0)
            return true;
        return searchHelp(word, root);
    }
};
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");