jeremymouzin
/ gist:085b976c806d55dd5735c6a6f0f4cfa8
Created
March 2, 2022 18:27
Exo CodinGame Benford Law
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| let total = arr.reduce((res,v) => res += v[1], 0); | |
| arr = arr.map(value => [value[0], value[1] / total * 100]); | |
| let benfordLaw = { | |
| '1': 30.1, | |
| '2': 17.6, | |
| '3': 12.5, | |
| '4': 9.7, | |
| '5': 7.9, | |
| '6': 6.7, | |
| '7': 5.8, |
jeremymouzin
/ tests.txt
Created
February 15, 2021 18:26
Dernier rapport de tests pour détecter les surfaces sur le site le bon coin
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| > lemeilleurcoin@1.0.0 test /Users/jeremy/Desktop/github/lemeilleurcoin | |
| > jest | |
| (node:3234) ExperimentalWarning: The fs.promises API is experimental | |
| PASS tests/commun.test.js (8.881 s) | |
| Conversion de surfaces | |
| ✓ convertir 7 m² en 7 m² (3 ms) | |
| ✓ convertir 7 M² en 7 m² | |
| ✓ convertir 7 M² en 7 m² |
jeremymouzin
/ tests.txt
Created
February 15, 2021 17:41
Rapport de tests sur la détection de surface dans une description
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| > lemeilleurcoin@1.0.0 test /Users/jeremy/Desktop/github/lemeilleurcoin | |
| > jest | |
| (node:1851) ExperimentalWarning: The fs.promises API is experimental | |
| PASS tests/commun.test.js (9.408 s) | |
| Conversion de surfaces | |
| ✓ convertir 7 m² en 7 m² (2 ms) | |
| ✓ convertir 7 M² en 7 m² | |
| ✓ convertir 7 M² en 7 m² |
jeremymouzin
/ lemeilleurcoin.js
Created
January 29, 2021 14:01
Script pour améliorer le site du bon coin pour mes besoins
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| // ==UserScript== | |
| // @name Le Meilleur Coin | |
| // @namespace http://tampermonkey.net/ | |
| // @version 1.0 | |
| // @description Make the bon coin a better place AH AH AH AH | |
| // @author Jérémy Mouzin | |
| // @match https://www.leboncoin.fr/ventes_immobilieres/* | |
| // @grant none | |
| // ==/UserScript== |
jeremymouzin
/ challenge10.js
Created
July 4, 2020 13:08
Challenge 10 du Challenge JavaScript 10 jours de Scrimba
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| // Version lisible pour les débutants | |
| function maxMultiple(diviseur, limite) { | |
| // On part de la limite et on teste chaque valeur | |
| for (let i = limite; i > 0; i--) { | |
| // Si i est divisible par diviseur, son reste de la | |
| // division euclidienne de i par diviseur sera 0 | |
| if (i % diviseur === 0) { | |
| return i; | |
| } | |
| } |
jeremymouzin
/ challenge9.js
Created
July 3, 2020 11:59
Challenge 9 du Challenge JavaScript 10 jours de Scrimba
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| // Version lisible pour les débutants | |
| function largestNumber(num) { | |
| // On concatène plusieurs 9 côte à côte | |
| const nombreChaine = '9'.repeat(num); | |
| // On convertit la chaîne en un nombre | |
| return Number(nombreChaine); | |
| } | |
| // Version compacte de la solution précédente | |
| function largestNumber(num) { |
jeremymouzin
/ challenge8.js
Created
July 2, 2020 21:14
Challenge 8 du Challenge JavaScript 10 jours de Scrimba
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| // Version lisible pour les débutants | |
| // Gère les espaces dans chaine | |
| function firstDigit(chaine) { | |
| for (caractere of chaine) { | |
| if ("0123456789".includes(caractere)) { | |
| return caractere; | |
| } | |
| } | |
| } |
jeremymouzin
/ challenge7.js
Created
July 1, 2020 13:03
Challenge 7 du Challenge JavaScript 10 jours de Scrimba
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| // Solution classique grâce à la récursivité | |
| function factorialNumber(num) { | |
| if (num !== 1) { | |
| return num * factorialNumber(num - 1); | |
| } else return 1; | |
| } | |
| // Solution sans récursivité | |
| function factorialNumber(num) { | |
| let resultat = 1; |
jeremymouzin
/ challenge6.js
Created
June 30, 2020 19:45
Challenge 6 du Challenge JavaScript 10 jours de Scrimba
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| // Solution simple via concaténation | |
| function encloseInBrackets(chaine) { | |
| return '(' + chaine + ')'; | |
| } | |
| // Solution simple via template strings | |
| function encloseInBrackets(chaine) { | |
| return `(${chaine})`; | |
| } |
jeremymouzin
/ challenge5.js
Created
June 29, 2020 13:55
Challenge 5 du Challenge JavaScript 10 jours de Scrimba
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| // Version lisible pour débutants | |
| function caseInsensitivePalindrome(mot) { | |
| // On met tout en minuscules pour rester insensible à la casse | |
| mot = mot.toLowerCase(); | |
| // On transforme la chaîne en tableau de caractères... | |
| let palindrome = mot.split(''); | |
| // ...pour pouvoir utiliser les méthodes des tableaux | |
| // et ainsi "retourner" facilement le mot | |
| palindrome = palindrome.reverse().join(''); | |
| // On teste si le mot retourné est égal au mot |
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