时间:2022-07-25 10:53:32 | 栏目:C代码 | 点击:次
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.
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:
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。
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参考资料:
https://leetcode.com/discuss/36246/my-java-trie-based-solution