(string 数组) leetcode 804. Unique Morse Code Words
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, "cba" 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
wordswill be at most100. - Each
words[i]will have length in range[1, 12]. words[i]will only consist of lowercase letters.
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
这个题说的是将字符串中的每个字符转换为相应的摩尔字符,然后拼接,最后判断有几个是不同的(删去重复的)。
这个题我首次尝试用了string 数组。
C++代码:
class Solution {
public:
int uniqueMorseRepresentations(vector<string>& words) {
set<string> s;
int len = words.size();
string s1[] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
for(int i = ; i < len; i++){
int j = words[i].length();
string str = "";
for(int k = ; k < j; k++){
str += s1[words[i][k] - 'a'];
}
s.insert(str);
}
return s.size();
}
};
(string 数组) leetcode 804. Unique Morse Code Words的更多相关文章
- LeetCode 804. Unique Morse Code Words (唯一摩尔斯密码词)
题目标签:String 题目给了我们 对应每一个 字母的 morse 密码,让我们从words 中 找出 有几个不同的 morse code 组合. 然后只要遍历 words,把每一个word 转换成 ...
- [LeetCode] 804. Unique Morse Code Words 独特的摩斯码单词
International Morse Code defines a standard encoding where each letter is mapped to a series of dots ...
- Leetcode 804. Unique Morse Code Words 莫尔斯电码重复问题
参考:https://blog.csdn.net/yuweiming70/article/details/79684433 题目描述: International Morse Code defines ...
- LeetCode - 804. Unique Morse Code Words
International Morse Code defines a standard encoding where each letter is mapped to a series of dots ...
- LeetCode 804 Unique Morse Code Words 解题报告
题目要求 International Morse Code defines a standard encoding where each letter is mapped to a series of ...
- [LeetCode] 804. Unique Morse Code Words_Easy tag: Hash Table
International Morse Code defines a standard encoding where each letter is mapped to a series of dots ...
- 804. Unique Morse Code Words - LeetCode
Question 804. Unique Morse Code Words [".-","-...","-.-.","-..&qu ...
- 【Leetcode_easy】804. Unique Morse Code Words
problem 804. Unique Morse Code Words solution1: class Solution { public: int uniqueMorseRepresentati ...
- 【Leetcode】804. Unique Morse Code Words
Unique Morse Code Words Description International Morse Code defines a standard encoding where each ...
随机推荐
- vue之v-for使用说明
demo.html <!DOCTYPE html> <html lang="en" xmlns:v-bind="http://www.w3.org/19 ...
- 洛谷 P2921 [USACO08DEC]在农场万圣节Trick or Treat on the Farm
题目描述 每年,在威斯康星州,奶牛们都会穿上衣服,收集农夫约翰在N(1<=N<=100,000)个牛棚隔间中留下的糖果,以此来庆祝美国秋天的万圣节. 由于牛棚不太大,FJ通过指定奶牛必须遵 ...
- mysql语句-DML语句
DML语句 DML是指对数据库中表记录的操作,主要包括数据的增删改查以及更新,下面依次介绍 首先创建一张表:: 表名:emp 字段:ename varchar(20),hiredate date ,s ...
- LOJ2250 [ZJOI2017] 仙人掌【树形DP】【DFS树】
题目分析: 不难注意到仙人掌边可以删掉.在森林中考虑树形DP. 题目中说边不能重复,但我们可以在结束后没覆盖的边覆盖一个重复边,不改变方案数. 接着将所有的边接到当前点,然后每两个方案可以任意拼接.然 ...
- Codeforces300 F. A Heap of Heaps
Codeforces题号:#300F 出处: Codeforces 主要算法:树状数组/线段树 难度:4.6 思路分析: 在没看到数据范围之前真是喜出望外,直到发现O(n^2)会被卡…… 其实也不是特 ...
- Python里的单下划线,双下划线,以及前后都带下划线的意义
Python里的单下划线,双下划线,以及前后都带下划线的意义: 单下划线如:_name 意思是:不能通过from modules import * 导入,如需导入需要:from modules imp ...
- C Looooops POJ - 2115 (exgcd)
一个编译器之谜:我们被给了一段C++语言风格的循环 for(int i=A;i!=B;i+=C) 内容; 其中所有数都是k位二进制数,即所有数时膜2^k意义下的.我们的目标时球出 内容 被执行了多少次 ...
- MT【297】任意四边形的一个向量性质
在平面四边形$ABCD$中,已知$E,F,G,H$分别是棱$AB,BC,CD,DA$的中点,若$|EG|^2-|HF|^2=1,$设$|AD|=x,|BC|=y,|AB|=z,|CD|=1,$则$\d ...
- 【LOJ6053】简单的函数(min_25筛)
题面 LOJ 题解 戳这里 #include<iostream> #include<cstdio> #include<cstdlib> #include<cs ...
- 【BZOJ3451】Normal (点分治)
[BZOJ3451]Normal (点分治) 题面 BZOJ 题解 显然考虑每个点的贡献.但是发现似乎怎么算都不好计算其在点分树上的深度. 那么考虑一下这个点在点分树中每一次被计算的情况,显然就是其在 ...