描述
International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on.
For convenience, the full table for the 26 letters of the English alphabet is given below:
[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
Now, given a list of words, each word can be written as a concatenation of the Morse code of each letter. For example, "cab" can be written as "-.-..--...", (which is the concatenation "-.-." + ".-" + "-..."). We'll call such a concatenation, the transformation of a word.
Return the number of different transformations among all words we have.
Example:
Input: words = ["gin", "zen", "gig", "msg"]
Output: 2
Explanation:
The transformation of each word is:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."
There are 2 different transformations, "--...-." and "--...--.".
Note:
- 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.
解析
根据题意，已经有了 26 个字母对应的摩斯码，那么只需要用字典类型的对象，在遍历每个 word 的时候，将 word 中的每个字符转换为摩斯码并连成字符串，最后将得到的所有字符串去重即可得到结果个数。
解答
class Solution(object):
def uniqueMorseRepresentations(self, words):
"""
:type words: List[str]
:rtype: int
"""
res = set()
morse = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
for word in words:
tmp = ""
for c in word:
tmp += morse[ord(c)-97]
res.add(tmp)
return len(res)
运行结果
Runtime: 20 ms, faster than 81.22% of Python online submissions for Unique Morse Code Words.
Memory Usage: 13.6 MB, less than 47.68% of Python online submissions for Unique Morse Code Words.
原题链接：leetcode.com/problems/un…
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