[leetcode-676-Implement Magic Dictionary]

Posted 2020-10-06 hellowOOOrld

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:

1. You may assume that all the inputs are consist of lowercase letters a-z.
2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
3. 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.

思路：

用一个map记录长度为len的所有字符串，当search的时候根据字符串的长度，找到所有相同字符串的长度，然后统计map里的字符串与要

查询的字符串字符差异个数。

也可以用一个set保存所有字符串，search的时候将字符串每一位都进行a-z的替换，随时查看是否在set里面即可。

 
 map<int,vector<string>>mp;
 
 MagicDictionary()
 {
        mp.clear();
 }
    
    /** Build a dictionary through a list of words */
    void buildDict(vector<string> dict) {
        if(dict.size()==0)return;
    for(auto s:dict)
    {
      mp[s.length()].push_back(s);
    }
    }
    
    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
    bool search(string word) {
        int len = word.length();
    int diff =0;
    if(mp[len].size()==0)return false;
    
    for(int i=0;i<mp[len].size();i++)
    {
      diff==0;
      for(int j =0;j<mp[len][i].length();j++)
      {
        if(word[j]!=mp[len][i][j])diff++;
      }
      if(diff==1 )return true;      
    }
    return false;
    }