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:

  1. [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

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.

  1. Example:
  2. Input: words = ["gin", "zen", "gig", "msg"]
  3. Output: 2
  4. Explanation: 
  5. The transformation of each word is:
  6. "gin" -> "--...-."
  7. "zen" -> "--...-."
  8. "gig" -> "--...--."
  9. "msg" -> "--...--."
  11. There are 2 different transformations, "--...-." and "--...--.".

Note:

  • The length of words will be at most 100.
  • 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

  1. class Solution {
  2.     public int uniqueMorseRepresentations(String[] words) {
  3.         Set<String> set = new HashSet<>();
  4.         String[] hash = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
  6.         for (String word : words) {
  7.             StringBuilder builder = new StringBuilder();
  8.             for (char c : word.toCharArray()) {
  9.                 builder.append(hash[c - 'a']);
  10.             }
  11.             set.add(builder.toString());
  12.         }
  14.         return set.size();
  15.     }
  16. }