[ LeetCode ] [ JavaScript ] - 1309. Decrypt String from Alphabet to Integer Mapping 難度 ★★ @ 西瓜是肥貓

先看看題目：

1309. Decrypt String from Alphabet to Integer Mapping

Easy

Given a string s formed by digits ('0' - '9') and '#' . We want to map s to English lowercase characters as follows:

Characters ('a' to 'i') are represented by ('1' to '9') respectively.
Characters ('j' to 'z') are represented by ('10#' to '26#') respectively.

Return the string formed after mapping.

It's guaranteed that a unique mapping will always exist.

Example 1:

Input: s = "10#11#12"
Output: "jkab"
Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

Example 2:

Input: s = "1326#"
Output: "acz"

Example 3:

Input: s = "25#"
Output: "y"

Example 4:

Input: s = "12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#"
Output: "abcdefghijklmnopqrstuvwxyz"

Constraints:

1 <= s.length <= 1000
s[i] only contains digits letters ('0'-'9') and '#' letter.
s will be valid string such that mapping is always possible.

附上通關證明：

截圖 2020-06-15 下午2.42.58

我的想法：題目是說把數字1~9跟10#~26#轉換成英文字母，其實很簡單，就從尾巴開始，看到#就取出倒數二三個然後切割，沒有#就取一個然後切割，然後根據題目去做數字轉字母。

試著寫成js就過了，不過速度跟記憶算是慢的，我看了一下網路有自己造map的可能會比較快，但我懶，使用了預設函數。

/**
 * @param {string} s
 * @return {string}
 */
var freqAlphabets = function(s) {
    result = ""
    while(s.length>0){
        lastchar = s.charAt(s.length-1)
        alphabet="";
        if(lastchar ==='#'){
            alphabet = numToalphabet(s.substring(s.length-3,s.length-1)) 
            s = s.substring(0,s.length-3)
        }else{
            alphabet = numToalphabet(s.substring(s.length-1,s.length))
            s = s.substring(0,s.length-1)
        }
        result = alphabet + result
    }
    return result
}
function numToalphabet(numString){
    return String.fromCharCode(parseInt(numString)+96)
}