LeetCode刷题记录——第804题（唯一莫尔斯密码词）

版权声明：此BLOG为个人BLOG，内容均来自原创及互连网转载。最终目的为记录自己需要的内容或自己的学习感悟，不涉及商业用途，转载请附上原博客。 https://blog.csdn.net/bulo1025/article/details/88351519

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 “–…--.”.

思路分析

• 遍历words中每个单词的每个字母
• 用ord函数，找到它该字母的ascii码，与a的ascii码相减就得到了它的索引
• 将dic中对应的morse码add到集合中，就不需要管重复的个数了

代码示例

class Solution(object):
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        dic = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        res = set()
        for word in words:
            p = ''
            for w in word:
                p += dic[ord(w) - ord('a')]
            res.add(p)
        return len(res)