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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:
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;
}
}