[leetcode-676-Implement Magic Dictionary]
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.
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()==)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 =;
if(mp[len].size()==)return false; for(int i=;i<mp[len].size();i++)
{
diff==;
for(int j =;j<mp[len][i].length();j++)
{
if(word[j]!=mp[len][i][j])diff++;
}
if(diff== )return true;
}
return false;
}
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