作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• * 题目描述：
  • 题目大意
  • 解题方法
  • * set + map
    • set + 字典
  • 日期

题目地址：https://leetcode.com/problems/unique-morse-code-words/description/

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.

找出一组字符串进行莫尔斯电码的编码有多少种不同情况。

set + map

找出多少种不同的情况，完全可以用len(set())的方式进行处理。所以，先得到每个字符的莫尔斯电码，然后把字符串所有进行拼接。很简单了哈。

时间复杂度是O(n)，空间复杂度是O(1)。

代码：

class Solution(object):
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        moorse = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        trans = lambda x: moorse[ord(x) - ord('a')]
        map_word = lambda word: ''.join([trans(x) for x in word])
        res = map(map_word, words)
        return len(set(res))

set + 字典

上面的做法需要ord函数，其实如果用字典速度会更快的。另外有个加速的经验就是，对字符串word+="dfs"操作是很慢的，改成word = word + "dfs"会变快很多。

时间复杂度是O(n)，空间复杂度是O(26)。

class Solution:
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        morse = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        english = ['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']
        edict = dict(zip(english, morse))
        res = set()
        for word in words:
            mword = ""
            for w in word:
                mword = mword + edict[w]
            res.add(mword)
        return len(res)

2018 年 3 月 31 日 ———— 晚上睡不好，一天没精神啊
2018 年 11 月 2 日 —— 浑浑噩噩的一天