C++实现LeetCode(211.添加和查找单词-数据结构设计)

篇文章主要介绍了C++实现LeetCode(211.添加和查找单词-数据结构设计),本篇文章通过简要的案例,讲解了该项技术的了解与使用,以下就是详细内容,需要的朋友可以参考下

[LeetCode] 211.Add and Search Word - Data structure design 添加和查找单词-数据结构设计

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.

click to show hint.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

LeetCode出新题的速度越来越快了，有点跟不上节奏的感觉了。这道题如果做过之前的那道 Implement Trie (Prefix Tree) 实现字典树(前缀树)的话就没有太大的难度了，还是要用到字典树的结构，唯一不同的地方就是search的函数需要重新写一下，因为这道题里面'.'可以代替任意字符，所以一旦有了'.'，就需要查找所有的子树，只要有一个返回true，整个search函数就返回true，典型的DFS的问题，其他部分跟上一道实现字典树没有太大区别，代码如下：

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52

class WordDictionary { public:     struct TrieNode {     public:         TrieNode *child[26];         bool isWord;         TrieNode() : isWord(false) {             for (auto &a : child) a = NULL;         }     };           WordDictionary() {         root = new TrieNode();     }           // Adds a word into the data structure.     void addWord(string word) {         TrieNode *p = root;         for (auto &a : word) {             int i = a - 'a';             if (!p->child[i]) p->child[i] = new TrieNode();             p = p->child[i];         }         p->isWord = 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 searchWord(word, root, 0);     }           bool searchWord(string &word, TrieNode *p, int i) {         if (i == word.size()) return p->isWord;         if (word[i] == '.') {             for (auto &a : p->child) {                 if (a && searchWord(word, a, i + 1)) return true;             }             return false;         } else {             return p->child[word[i] - 'a'] && searchWord(word, p->child[word[i] - 'a'], i + 1);         }     }       private:     TrieNode *root; }; // Your WordDictionary object will be instantiated and called as such: // WordDictionary wordDictionary; // wordDictionary.addWord("word"); // wordDictionary.search("pattern");

讨论：这道题有个很好的Follow up，就是当搜索的单词中存在www.meimeitu8.com星号怎么搞，星号的定义和Wildcard Matching中一样，可以代表任意的字符串，包括空字符串，请参见评论区1楼。

类似题目：

Implement Trie (Prefix Tree)

Wildcard Matching

参考资料：

https://leetcode.com/discuss/36246/my-java-trie-based-solution

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