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, "cba" 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.---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
这个题说的是将字符串中的每个字符转换为相应的摩尔字符，然后拼接，最后判断有几个是不同的（删去重复的）。
这个题我首次尝试用了string 数组。
C++代码：
class Solution { public: int uniqueMorseRepresentations(vector<string>& words) { set<string> s; int len = words.size(); string s1[26] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."}; for(int i = 0; i < len; i++){ int j = words[i].length(); string str = ""; for(int k = 0; k < j; k++){ str += s1[words[i][k] - 'a']; } s.insert(str); } return s.size(); } };
转载于:https://www.cnblogs.com/Weixu-Liu/p/10693952.html
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