one liner second creative solution for [old] Digits Multiplication by Sim0000 - javascript coding challenges

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one liner second solution in Creative category for [old] Digits Multiplication by Sim0000

var digitsMultip=n=>!n||digitsMultip(n/10^0)*(n%10||1)

Sept. 15, 2016

Comments:

sweatC on Jan. 27, 2017, 1:37 p.m.
Wow! It is really magic, can you tell me how it works?

Sim0000 on Jan. 27, 2017, 2:37 p.m.
Basically, this code uses recursive calls, and it processes in order from the lower digit. For example, we calculate in this order: <pre class='brush: javascript'>f(123) = f(12) * 3 = f(1) * 2 * 3 = f(0) * 1 * 2 * 3 = 1 * 1 * 2 * 3 </pre> It is a rewritten code for easy understanding. <pre class='brush: javascript'>function digitsMultip(n) { if(n == 0){ // !n || return 1; // } else { // let h = Math.floor(n / 10) // = n/10^0 let k = n % 10 != 0 ? n % 10 : 1 // = n%10||1 (lowest digit) return digitsMultip(h) * k } } </pre> Have a nice coding :-)

sweatC on Jan. 27, 2017, 4:14 p.m.
I understood, I was very confused when I first looked at this code Thank you for your feedback =)

veghadam1991 on Sept. 25, 2017, 5:23 p.m.
I love that! :))

Sim0000 on May 29, 2018, 5:51 a.m.
I can write more short :-) <pre class='brush: javascript'>var f=n=>!n||f(n/10^0)*(n%10||1),digitsMultip=f </pre>

Tairraos on March 13, 2019, 1:53 a.m.
more short: var f=digitsMultip=n=>!n||f(n/10^0)*(n%10||1)

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