标签:win lse char 实现 搜索 bad bool returns children
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; } };
LeetCode-211 Add and Search Word - Data structure design
标签:win lse char 实现 搜索 bad bool returns children
原文地址:https://www.cnblogs.com/heyn1/p/11050365.html