C++实现LeetCode(676.实现神奇字典)

[LeetCode] 676.Implement Magic Dictionary 实现神奇字典

Implement a magic directory with buildDict, and search methods.

For the method buildDict, you'll be given a list of non-repetitive words to build a dictionary.

For the method search, you'll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:

Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False

Note:

You may assume that all the inputs are consist of lowercase letters a-z.
For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.

这道题让我们设计一种神奇字典的数据结构，里面有一些单词，实现的功能是当我们搜索一个单词，只有存在和这个单词只有一个位置上的字符不相同的才能返回true，否则就返回false，注意完全相同也是返回false，必须要有一个字符不同。博主首先想到了One Edit Distance那道题，只不过这道题的两个单词之间长度必须相等。所以只需检测和要搜索单词长度一样的单词即可，所以我们用的数据结构就是根据单词的长度来分，把长度相同相同的单词放到一起，这样就可以减少搜索量。那么对于和要搜索单词进行比较的单词，由于已经保证了长度相等，我们直接进行逐个字符比较即可，用cnt表示不同字符的个数，初始化为0。如果当前遍历到的字符相等，则continue；如果当前遍历到的字符不相同，并且此时cnt已经为1了，则break，否则cnt就自增1。退出循环后，我们检测是否所有字符都比较完了且cnt为1，是的话则返回true，否则就是跟下一个词比较。如果所有词都比较完了，则返回false，参见代码如下：

解法一：

				?

									class MagicDictionary {

									public:

									    /** Initialize your data structure here. */

									    MagicDictionary() {}

									    /** Build a dictionary through a list of words */

									    void buildDict(vector<string> dict) {

									        for (string word : dict) {

									            m[word.size()].push_back(word);

									        }

									    }

									    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */

									    bool search(string word) {

									        for (string str : m[word.size()]) {

									            int cnt = 0, i = 0;

									            for (; i < word.size(); ++i) {

									                if (word[i] == str[i]) continue;

									                if (word[i] != str[i] && cnt == 1) break; 

									                ++cnt;

									            }

									            if (i == word.size() && cnt == 1) return true;

									        }

									        return false;

									    }

									private:

									    unordered_map<int, vector<string>> m;

									};

下面这种解法实际上是用到了前缀树中的search的思路，但是我们又没有整个用到prefix tree，博主感觉那样写法略复杂，其实我们只需要借鉴一下search方法就行了。我们首先将所有的单词都放到一个集合中，然后在search函数中，我们遍历要搜索的单词的每个字符，然后把每个字符都用a-z中的字符替换一下，形成一个新词，当然遇到本身要跳过。然后在集合中看是否存在，存在的话就返回true。记得换完一圈字符后要换回去，不然就不满足只改变一个字符的条件了，参见代码如下：

解法二：

				?

									class MagicDictionary {

									public:

									    /** Initialize your data structure here. */

									    MagicDictionary() {}

									    /** Build a dictionary through a list of words */

									    void buildDict(vector<string> dict) {

									        for (string word : dict) s.insert(word);

									    }

									    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */

									    bool search(string word) {

									        for (int i = 0; i < word.size(); ++i) {

									            char t = word[i];

									            for (char c = 'a'; c <= 'z'; ++c) {

									                if (c == t) continue;

									                word[i] = c;

									                if (s.count(word)) return true;

									            }

									            word[i] = t;

									        }

									        return false;

									    }

									private:

									    unordered_set<string> s;

									};