Implement Magic Dictionary问题及解法

问题描述：

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.

示例:

Input: buildDict(["hello", "leetcode"]), Output: NullInput: search("hello"), Output: FalseInput: search("hhllo"), Output: TrueInput: search("hell"), Output: FalseInput: search("leetcoded"), Output: False

问题分析：

问题要求我们只改变字串中一个字符，其他的并保持原样，那我们可以考虑将剩余的字符组成一个新串，作为一个key值，将改变的字符的索引和字符值保存为pair，只要我们就构成了对应关系。在搜索时，我们只要找到key值相同的，且索引值也相同，但字符值不同的结果即可。

过程详见代码：

class MagicDictionary {public:unordered_map<string, vector<pair<int, char>>> map;/** Initialize your data structure here. */MagicDictionary() {}/** Build a dictionary through a list of words */void buildDict(vector<string> dict) {for (auto s : dict){for (int j = 0; j < s.length(); j++){string key = s.substr(0, j) + s.substr(j + 1);map[key].emplace_back(pair<int, char>(j, s[j]));}}}/** 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.length(); i++){string key = word.substr(0, i) + word.substr(i + 1);if (map.count(key)){for (auto res : map[key]){if (res.first == i && res.second != word[i]) return true;}}}return false;}};/** * Your MagicDictionary object will be instantiated and called as such: * MagicDictionary obj = new MagicDictionary(); * obj.buildDict(dict); * bool param_2 = obj.search(word); */