Persistence

문제 Compare two arrays and return a new array with any items only found in one of the two given arrays, but not both. 예시 diffArray(["andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]) // should return ["diorite", "pink wool"]. 내가 푼 답 function diffArray(arr1, arr2) { return [...arr1.filter(n => arr2.indexOf(n) === -1), ...arr2.filter(n =..

문제 We'll pass you an array of two numbers. Return the sum of those two numbers plus the sum of all the numbers between them. The lowest number will not always come first. 예시 sumAll([1, 4]) // should return 10. sumAll([10, 5]) // should return 45. 내가 푼 답 function sumAll(arr) { var sorted = [].concat(arr).sort((a,b) => a-b); var result = 0; for(var i = sorted[0] ; i

문제 You will be provided with an initial array (the first argument in the destroyer function), followed by one or more arguments. Remove all elements from the initial array that are of the same value as these arguments. 예시 destroyer([1, 2, 3, 1, 2, 3], 2, 3) should return [1, 1]. destroyer([1, 2, 3, 5, 1, 2, 3], 2, 3) should return [1, 5, 1]. destroyer([3, 5, 1, 2, 2], 2, 3, 5) should return [1]...