Description
Implement a magic dictionary 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.
分析
题目的意思是:实现一个magic dictionary的功能,要实现search,buildDict两大函数。
- 用了一个hash表,用单词的size当作key,单词当作值,设计不错。
- 想到了就很简单,我也做不出来,学习一下,努力加油。
代码
class MagicDictionary {
private:
unordered_map<int,vector<string>> m;
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(str[i]!=word[i]){
cnt++;
if(cnt>1){
break;
}
}
}
if(i==word.size()&&cnt==1){
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);
*/