LWC 49：676. Implement Magic Dictionary

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;    }}