标签：most 多少 control 判断 bsp word ons 返回 elf
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:
words will be at most 100.words[i] will have length in range [1, 12].words[i] will only consist of lowercase letters.给定了26字母的摩斯电码的编码，给一组单词，把每个单词都转成摩斯码，返回有多少个不同的摩斯码。
解法：题目很简单，直接转换判断即可。
Java:
public int uniqueMorseRepresentations(String[] words) {
String[] d = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
HashSet<String> s = new HashSet<>();
for (String word : words) {
String code = "";
for (char c : word.toCharArray()) code += d[c - ‘a‘];
s.add(code);
}
return s.size();
}
Python:
def uniqueMorseRepresentations(self, words):
d = [".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--",
"-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."]
return len({‘‘.join(d[ord(i) - ord(‘a‘)] for i in w) for w in words})
Python:
# Time: O(n), n is the sume of all word lengths
# Space: O(n)
class Solution(object):
def uniqueMorseRepresentations(self, words):
"""
:type words: List[str]
:rtype: int
"""
MORSE = [".-", "-...", "-.-.", "-..", ".", "..-.", "--.",
"....", "..", ".---", "-.-", ".-..", "--", "-.",
"---", ".--.", "--.-", ".-.", "...", "-", "..-",
"...-", ".--", "-..-", "-.--", "--.."]
lookup = {"".join(MORSE[ord(c) - ord(‘a‘)] for c in word) for word in words}
return len(lookup)
Python: wo
class Solution(object):
def uniqueMorseRepresentations(self, words):
"""
:type words: List[str]
:rtype: int
"""
m = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
trans = []
res = 0
for word in words:
temp = ‘‘
for c in word:
temp += m[ord(c) - 97]
if temp not in trans:
trans.append(temp)
res += 1
return res
C++:
int uniqueMorseRepresentations(vector<string>& words) {
vector<string> d = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
unordered_set<string> s;
for (auto word : words) {
string code;
for (auto c : word) code += d[c - ‘a‘];
s.insert(code);
}
return s.size();
}
C++:
class Solution {
public:
int uniqueMorseRepresentations(vector<string>& words) {
vector<string> morse{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
unordered_set<string> s;
for (string word : words) {
string t = "";
for (char c : word) t += morse[c - ‘a‘];
s.insert(t);
}
return s.size();
}
};
假定followup: 给一个单词的摩斯码，问有几种可能的单词，比如："--...-."，至少有两种zen和gin
[LeetCode] 804. Unique Morse Code Words 独特的摩斯码单词
标签：most 多少 control 判断 bsp word ons 返回 elf
原文地址：https://www.cnblogs.com/lightwindy/p/9847603.html
