LWC 49：676. Implement Magic Dictionary
传送门：676. Implement Magic Dictionary
Problem:
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.
无脑做法：
用MAP做长度的一个映射，把长度相同的放入一个LIST中，接着判断LIST中是否存在编辑距离为1的单词。
代码如下：
class MagicDictionary {
/** Initialize your data structure here. */
Map<Integer, List<String>> map;
public MagicDictionary() {
map = new HashMap<>();
}
/** Build a dictionary through a list of words */
public void buildDict(String[] dict) {
for (String word : dict) {
map.computeIfAbsent(word.length(), k -> new ArrayList<>()).add(word);
}
}
/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
public boolean search(String word) {
int len = word.length();
if (!map.containsKey(len)) return false;
for (String dict : map.get(word.length())) {
if (valid(dict, word)) return true;
}
return false;
}
public boolean valid(String word, String dict) {
int n = word.length();
int j = 0;
for (int i = 0; i < n; ++i) {
if (word.charAt(i) == dict.charAt(i)) j++;
}
return j == n - 1;
}
}
