Unique Morse Code Words (E)
题目
International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a"
maps to ".-"
, "b"
maps to "-..."
, "c"
maps to "-.-."
, and so on.
For convenience, the full table for the 26 letters of the English alphabet is given below:
[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
Now, given a list of words, each word can be written as a concatenation of the Morse code of each letter. For example, "cab" can be written as "-.-..--...", (which is the concatenation "-.-." + ".-" + "-...
"). We'll call such a concatenation, the transformation of a word.
Return the number of different transformations among all words we have.
Example:
Input: words = ["gin", "zen", "gig", "msg"]
Output: 2
Explanation:
The transformation of each word is:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."
There are 2 different transformations, "--...-." and "--...--.".
Note:
- The length of
words
will be at most100
. - Each
words[i]
will have length in range[1, 12]
. words[i]
will only consist of lowercase letters.
题意
将一连串单词转化为摩斯电码,并统计不相同的电码的个数。
思路
HashSet解决。
代码实现
Java
class Solution {
public int uniqueMorseRepresentations(String[] words) {
Set<String> set = new HashSet<>();
String[] hash = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
for (String word : words) {
StringBuilder builder = new StringBuilder();
for (char c : word.toCharArray()) {
builder.append(hash[c - 'a']);
}
set.add(builder.toString());
}
return set.size();
}
}