本篇為LeetCode上演算法的簡單問題，Unique Morse Code Words。
題目如下：
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 1: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 "--...--.".Notes:
- 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.
摩斯密碼定義一套標準的編碼，每一個英文字母對照著一組由點(dot)'.'和線(dash)'-'組合成的符號，例如字母"a"對照的是".-"，"b"對照的是"-..."，"c"對照的是"-.-."。
全部的字母定義如上方區塊中的陣列。
輸入參數為多個文字組成的列表，列表中每個文字的字母可轉成摩斯密碼轉成的符號，例如"cba"可轉譯成"-.-..--..."（由"-.-." + "-..." + ".-"串接成）。
計算每個文字所轉譯的摩斯密碼串接符號共有幾組不同。
Notes:
- 文字列表的個數不超過100個字。
- 每個文字的字數為1到12。
- 每個文字皆為小寫英文字母。
解法的基本步驟為：
- 迴圈文字列表。
- 將每個文字轉成字符陣列並迴圈
- 將每個字符依照ASCII表的規則扣除97（直接減
'a'也可），即為摩斯密碼陣列中該字母所對映的符號索引。 - 從摩斯密碼陣列表中取出該字母對映的密碼，並串接成字串。
- 將字串放入Set中確保唯一。
- 計算Set中元素的數量並回傳。
public static int uniqueMorseRepresentations(String[] words) {
String[] codes = { ".-", "-...", "-.-.", "-..", ".", "..-.",
"--.", "....", "..", ".---", "-.-", ".-..", "--",
"-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-",
"...-", ".--", "-..-", "-.--", "--.." };
Set<String> set = new HashSet<>();
for (String word : words) {
StringBuilder sb = new StringBuilder(12);
for(char c : word.toCharArray()) {
sb.append(codes[(c - 97)]); // 或codes[(c - 'a')]
}
set.add(sb.toString());
}
return set.size();
}
在以StringBuilder串接符號時，依題目條件可以先宣告好空間(capacity)為12。（StringBuilder預設空間長度為16，）
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