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 most100
. - Each
words[i]
will have length in range[1, 12]
. words[i]
will only consist of lowercase letters.
class Solution { private String[] morseCodes = new String[]{ ".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."}; public int uniqueMorseRepresentations(String[] words) { if (words == null || words.length == 0) { return 0; } Set<Integer> set = new HashSet<>(); int ans = 0; for (String word : words) { int code = getHashValue(word); if (!set.contains(code)) { ans++; set.add(code); } } return ans; } private int getHashValue(String s) { int ans = 0; StringBuilder sb = new StringBuilder(); for (int i = 0; i < s.length(); i++) { sb.append(morseCodes[s.charAt(i) - 'a']); } ans = morseToNumber(sb.toString()); return ans; } private int morseToNumber(String s) { int ans = 0; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); ans = ans * 2 + (c == '-' ? 1 : 0); } return ans; } }
No comments:
Post a Comment