1013. 独特的摩尔斯编码

1013. 独特的摩尔斯编码

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摩尔斯电码定义了一种标准编码，把每个字母映射到一系列点和短划线，例如：a -> .-，b -> -...，c ->-.-.。

给出26个字母的完整编码表格：

[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

现在给定一个单词列表，每个单词中每个字母可以写成摩尔斯编码。 例如，cab可以写成-.-.-....-，（把c,a,b的莫尔斯编码串接起来）。 我们称之为一个词的转换。

返回所有单词中不同变换的数量。

样例

样例1：

输入: words = ["gin", "zen", "gig", "msg"]
输出: 2
解释: 
每一个单词的变换是:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."

这里有两种不同的变换结果： "--...-."和"--...--.".

样例2：

输入: words = ["a", "b"]
输出: 2
解释: 
每一个单词的变换是:
"a" -> ".-"
"b" -> "-..."
这里有两种不同的变换结果：".-" and "-...".

注意事项

• 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.

输入测试数据 (每行一个参数)如何理解测试数据？

class Solution:
    """
    @param words: the given list of words
    @return: the number of different transformations among all words we have
    """
    '''
    大致思路:
    1.初始化count = 0(计数，计算出不同的摩斯密码的个数),res = []。需要给出与摩斯密码映射字典
    2.循环words，s_res = '',每个字符串进行循环拼接s_res,如果不存在res里面的话，则append进来，最终返回res的长度即可
    '''
    def uniqueMorseRepresentations(self,words):
        count,res = 0,[]
        morse_dic = self.return_morse_dic()

        for column in words:
            s_res = ''
            for i in column:
                s_res += morse_dic[i]
            if s_res not in res:
                res.append(s_res)
        return len(res)
    
    def return_morse_dic(self):
        a = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        s = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
        dic = {}
        count = 0
        for i in s:
            dic[i] = a[count]
            count += 1
        return dic

 

来源：https://www.cnblogs.com/yunxintryyoubest/p/12602719.html