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 most 100.
Each words[i] will have length in range [1, 12].
words[i] will only consist of lowercase letters.
代码
运用Set集合去掉重复
class Solution {public int uniqueMorseRepresentations(String[] words) {String[] Morse={
".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};int len = words.length;Set<String> set = new HashSet<String>();for(int i = 0 ;i< len;i++){StringBuilder bu = new StringBuilder();String word = words[i];int wordLen = word.length();for(int j = 0;j<wordLen;j++){bu.append(Morse[word.charAt(j)-'a']);}set.add(bu.toString());}return set.size();}
}