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.

 

题目大意

实现两个操作：插入单词、查找单词（查找单词时的输入只能为 'a-z' 和 '.' ，其中 '.' 可以代表任何小写字母）。

 

示例

E

 

解题思路

解题思路类似于LeetCode - 208 Implement Trie，设置一个node class用来记录保存树结点中的节点信息。

插入单词是建树的过程，查找单词是搜索树的过程。

 

复杂度分析

时间复杂度：O(N)

空间复杂度：O(N)

 

代码

class TrieNode {public:    //该节点代表的字母是否是某个单词的最后一个字母    bool word;    TrieNode* children[26];    TrieNode() {        word = false;        memset(children, NULL, sizeof(children));    }};class WordDictionary {public:    /** Initialize your data structure here. */    WordDictionary() {            }        //建立一棵树    /** Adds a word into the data structure. */    void addWord(string word) {        TrieNode* node = root;        for (char c : word) {            if (!node -> children[c - 'a']) {                node -> children[c - 'a'] = new TrieNode();            }            node = node -> children[c - 'a'];        }        node -> word = true;    }        /** 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(word.c_str(), root);    }private:    TrieNode* root = new TrieNode();        bool search(const char* word, TrieNode* node) {        for (int i = 0; word[i] && node; i++) {            //若搜索的单词该位置不是‘.’，则直接进行比较，否则对所有子结点进行遍历搜索            if (word[i] != '.') {                node = node -> children[word[i] - 'a'];            } else {                TrieNode* tmp = node;                for (int j = 0; j < 26; j++) {                    node = tmp -> children[j];                    if (search(word + i + 1, node)) {                        return true;                    }                }            }        }        return node && node -> word;    }};