Python每日一编程小练习（2019.05.30）——唯一摩尔斯密码词

题目：

        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。

分析：

难点在于如何计算不重复翻译的数量，本题的解决办法参考了：https://blog.youkuaiyun.com/tanlangqie/article/details/78755665 

set()函数可以将列表中的元素无重复的抽取出来。

def mosimima(a):
    qq = list(str('abcdefghijklmnopqrstuvwxyz'))  #分割字符串
    mima = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
    h = []
    for i in range(0,len(a)):
        b = list(str(a[i]))
        d = []
        for j in range(0,len(b)):
            d.append(mima[qq.index(b[j])])       #拼接字符串
        h.append(''.join(d))
        #set()可将列表中的元素无重复的抽取出来，赋值给另一个列表
    return(len(set(h)))

运行结果：

注：本题目来源于：https://mp.weixin.qq.com/s/vbP4MSYUxx58cI5k1jxjrA

本题的难点解决办法参考了：https://blog.youkuaiyun.com/tanlangqie/article/details/78755665 

本题由笔者自己思考并编写代码实现，如有错误，欢迎指出。