Description
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.
题解
要求算出，将多个不同的字符串通过莫斯编码后，有多少种不同的结果。摩斯码与二进制类似，可通过0与1来存储，题目要求的单词长度最大为12，莫斯编码后的单个字符长度最大为4，因此需要的最大存储位数为48，因此使用long类型存储编码结果。
--...-. = 1100010
用集合保存编码结果，遍历字符串进行编码，每出现一个新的结果则将结果添加进集合内，并计数加一。