---
description: 可以先把題目存進 obj 或 array ，之後直接用 Big O (1) 抓就少一道轉的手續
---

# # 804 Unique Morse Code Words

[題目連結在此](https://leetcode.com/problems/contains-duplicate/)

```
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.

input: 給一字串陣列
output: 轉成 Morse Code 後回傳有幾個不同的結合
```

```
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.
*/

/**
 * @param {string[]} words
 * @return {number}
 */
var uniqueMorseRepresentations = function(words) {
  
};
```

### 怎麼解

前面會用 Array method，然後也利用 Set 回傳不重覆值的 size

* 先把英文數字轉成 a = 0, b = 1,... z = 26，這樣到時候才能抓摩斯密碼
* 看每一個 words (ex. "gin")
* 再看每一個 word (ex. g, i, n)
* 轉成摩斯密碼後合併起來

![](https://1787585077-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LqirD3iAZIDk4oDCfwS%2F-LqslLvJfjRZhvzytpmS%2F-LqsoEjKObNVgzTWdneg%2F2.jpg?alt=media\&token=6fe9242d-d1e1-4edb-9544-cd6cf93cb206)

```
const alphabet = {
    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: '--..' 
}

const uniqueMorseRepresentations = words => {  
    return new Set(words.map(word => {
        return word.split('').map(letter => alphabet[letter]).join(''))).size
    }
}
```

### 學到什麼?

* 本來一開始想法是用 charCodeAt 先計算每個字母代表數字然後再去對應 Morse Code

```
let getIndex = char => char.charCodeAt(0) - "a".charCodeAt(0)
```

結果其實可以一開始就存在 obj 裡，就不用再計算一次了．效能也增加 30% 呢
