本文共 1719 字，大约阅读时间需要 5 分钟。

题目描述：

国际摩尔斯密码定义一种标准编码方式，将每个字母对应于一个由一系列点和短线组成的字符串， 比如: “a” 对应 “.-”, “b” 对应 “-…”, “c” 对应 “-.-.”, 等等。

为了方便，所有26个英文字母对应摩尔斯密码表如下：

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

给定一个单词列表，每个单词可以写成每个字母对应摩尔斯密码的组合。例如，“cab” 可以写成 “-.-…–…”，(即 “-.-.” + “.-” + “-…” 字符串的结合)。我们将这样一个连接过程称作单词翻译。

返回我们可以获得所有词不同单词翻译的数量。

例如:

输入: words = [“gin”, “zen”, “gig”, “msg”] 输出: 2 解释: 各单词翻译如下: “gin” -> “–…-.” “zen” -> “–…-.” “gig” -> “–…--.” “msg” -> “–…--.”

共有 2 种不同翻译, “–…-.” 和 “–…--.”.

注意:

单词列表words 的长度不会超过 100。每个单词 words[i]的长度范围为 [1, 12]。每个单词 words[i]只包含小写字母。

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/unique-morse-code-words 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。 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])){     //若摩尔斯密码存在则退出                flag = 1;                break;            }        }        if (0 == flag){               strcpy(result[count], tmp[i]);      //若摩尔斯密码不存在则保存至result            count++;        }    }    return count;}

运行结果：