参考:https://blog.csdn.net/yuweiming70/article/details/79684433
题目描述:
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 most100
. - Each
words[i]
will have length in range[1, 12]
. words[i]
will only consist of lowercase letters.
翻译:摩尔斯电码是使用.-来表示字母,如果直接将字符串转换成摩尔斯电码,而没有空格的话,那么不同的字符串可能有相同的摩尔斯电码,下面给出一系列字符串,给出这些字符串能够得到多少种不同类的摩尔斯电码。
思路:
1.首先根据字母拼接对应的摩尔斯电码,然后将这一段电码做hash映射,可以看做是一段01串,直接转化成二进制即可(可能溢出int范围,但是没有关系,相同的字符串溢出后也是一样的)
2.将一系列字符串对应的hash值排序,去重即可。
代码:
class Solution { public: vector<string> mos={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."}; string to_mos(string &word) { string ans=""; for(int i=0;i<word.size();i++) ans += mos[word[i]-‘a‘]; return ans; } int removeDuplicates(vector<int>& nums) { int n=nums.size(); if(n < 2) return n; int id = 1; for(int i = 1; i < n; i++) if(nums[i] != nums[i-1]) nums[id++] =nums[i]; return id; } int hash(string &m) { int ans=0; for(int i=0;i<m.size();i++) { if(m[i]==‘-‘) ans++; ans*=2; } return ans; } int uniqueMorseRepresentations(vector<string>& words) { int n=words.size(); vector<string> mos_words(n); for(int i=0;i<n;i++) mos_words[i] = to_mos(words[i]); vector<int> mos_words_hash(n,0); for(int i=0;i<n;i++) mos_words_hash[i] = hash(mos_words[i]); sort(mos_words_hash.begin(),mos_words_hash.end()); return removeDuplicates(mos_words_hash); } };