【Leetcode -796.旋转字符串 -804.唯一摩尔斯密码词】

		bool rotateString(char* s, char* goal)
		{
		    //两个字符串如果相同，返回true
		    if (!strcmp(s, goal))
		        return true;
		
		    //先遍历s字符串，i是旋转的次数
		    for (int i = 0; i < strlen(s) - 1; i++)
		    {
		        //j挪动字符，temp 每次存放字符串的第一个字符，然后将后面的字符往前挪
		        int j = 0;
		        char temp = s[0];
		
		        //挪动字符
		        while (j < strlen(s))
		        {
		            char tmp = s[j];
		            s[j] = s[j + 1];
		            j++;
		        }
		
		        //将temp放到最后一个字符的位置
		        s[j - 1] = temp;
		
		        //判断旋转 i 次后和 goal 字符串是否相同，相同则返回true
		        if (!strcmp(s, goal))
		            return true;
		    }
		
		    //没有返回true则返回false
		    return false;
		}

		int uniqueMorseRepresentations(char** words, int wordsSize)
		{
		    //一共26种摩尔斯密码，每种摩尔斯密码最多5个字符
		    char temp[26][5] = { ".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.." };
		
		    //morse 数组存放不同单词的翻译；count统计不同单词翻译的数量
		    char morse[100][60] = { 0 };
		    int count = 0;
		    for (int i = 0; i < wordsSize; i++)
		    {
		        int flag = 0;
		
		        //tmp 数组存放每个字符串的摩尔斯密码；因为每个字符串的长度最长为 12 ，即最多有12个字母，转化成摩尔斯密码的长度就是60
		        char tmp[60] = { 0 };
		
		        //将摩尔斯密码对应的字符串追加到tmp数组中
		        for (int j = 0; j < strlen(words[i]); j++)
		        {
		            strcat(tmp, temp[words[i][j] - 'a']);
		        }
		
		        //比较morse数组和tmp数组，判断 tmp 是否已经在morse数组中 ，
		        //已经出现过则将 flag 置1，跳出循环
		        //否则将它放入 morse 数组中，并统计不同单词翻译的数量count
		        for (int k = 0; k < count; k++)
		        {
		            if (!strcmp(morse[k], tmp))
		            {
		                flag = 1;
		                break;
		            }
		        }
		
		        if (!flag)
		        {
		            strcpy(morse[count], tmp);
		            count++;
		        }
		    }
		    return count;
		}