技術標籤：LeetCode

題目描述：
國際摩爾斯密碼定義一種標準編碼方式，將每個字母對應於一個由一系列點和短線組成的字串， 比如: “a” 對應 “.-”, “b” 對應 “-…”, “c” 對應 “-.-.”, 等等。

為了方便，所有26個英文字母對應摩爾斯密碼錶如下：

[".-","-…","-.-.","-…",".","…-.","–.","…","…",".—","-.-",".-…","–","-.","—",".–.","–.-",".-.","…","-","…-","…-",".–","-…-","-.–","–…"]

給定一個單詞列表，每個單詞可以寫成每個字母對應摩爾斯密碼的組合。例如，“cab” 可以寫成 “-.-…–…”，(即 “-.-.” + “.-” + “-…” 字串的結合)。我們將這樣一個連線過程稱作單詞翻譯。

返回我們可以獲得所有詞不同單詞翻譯的數量。

例如:
輸入: words = [“gin”, “zen”, “gig”, “msg”]
輸出: 2
解釋:
各單詞翻譯如下:
“gin” -> “–…-.”
“zen” -> “–…-.”
“gig” -> “–…--.”
“msg” -> “–…--.”

共有 2 種不同翻譯, “–…-.” 和 “–…--.”.

注意:

單詞列表words 的長度不會超過 100。
每個單詞 words[i]的長度範圍為 [1, 12]。
每個單詞 words[i]只包含小寫字母。

來源：力扣（LeetCode）
連結：https://leetcode-cn.com/problems/unique-morse-code-words
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解答：

int uniqueMorseRepresentations(char ** words, int wordsSize){
    int i = 0;
    int j = 0;
    int n = 0;
    int k =
 
 0;
    int count = 0;
    char tmp[100][100] = {0};
    char result[120][100] = {0};
    int flag = 0;

    char res[26][10] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};

    for(i = 0; i < wordsSize;i++){
        j = 0;   
        flag = 0;

        while(words[i][j] != '\0'){ 
            n = words[i][j] - 'a';
            strcat(tmp[i], res[n]);             //每一行拼接到tmp[i]中
            
            j++;
        }

        for(k = 0; k < count; k++) {
            if(0 == strcmp(result[k], tmp[i])){  //若摩爾斯密碼存在則退出
                flag = 1;
                break;
            }
        }

        if (0 == flag){
            strcpy(result[count], tmp[i]);      //若摩爾斯密碼不存在則儲存至result
            count++;
        }
    }
    return count;
}

執行結果：