LWC 49：676. Implement Magic Dictionary

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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;
    }

}