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: NullInput: search("hello"), Output: FalseInput: search("hhllo"), Output: TrueInput: search("hell"), Output: FalseInput: 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 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 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 > 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 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 s;};
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