Javascript挑战：循环 - 多个条件 - 卡住，无法弄清楚

Question

我做了这个关于函数和执行上下文的模块-所有的问题都很顺利，但有一个挑战我花了很多时间，仍然无法解决。任何帮助都将不胜感激。谢谢

挑战问题说：写一个函数addingAllTheWeirdStuff，它将array2中所有奇数的和与array1中10以下的每个元素相加。

类似地，addingAllTheWeirdStuff还应该将array2中所有偶数的和与array1中超过10的元素相加。

BONUS：如果array2中的任何元素大于20，则在array1中的每个元素上加1。

// Uncomment these to check your work!
// console.log(addingAllTheWeirdStuff([1, 3, 5, 17, 15], [1, 2, 3, 4, 5])); // expected log [10, 12, 14, 23, 21]
// console.log(addingAllTheWeirdStuff([1, 3, 5, 17, 15, 1], [1, 2, 3, 4, 5, 22])); // expected log [11, 13, 15, 46, 44, 11]
// my attempt so far:
function addingAllTheWeirdStuff(array1, array2) {
// ADD CODE HERE
let result = []
for (let i = 0; i < array2.length; i++) {
if (array2[i] > 20) {
result = array1[i] += 1
}
}
for (let i = 0; i < array2.length; i++) {
if (array2[i] % 2 === 0 && array1[i] > 10) {
result = array1[i] + array2[i]
}
}
for (let i = 0; i < array2.length; i++) {
if (array2[i] % 2 !== 0 && array1[i] < 10) {
result = array1[i] + array2[i]
}
}
return result
}

Answer 1

使用reduce和map数组方法，使用三元运算符，可以轻松实现这一点

const array1 = [1, 3, 5, 17, 15];
const array2 = [1, 2, 3, 4, 5];
function addingAllTheWeirdStuff(array1, array2) {
const oddSum = array2.reduce((sum, current) => current % 2 ? current + sum : 0 + sum, 0)
const oddEven = array2.reduce((sum, current) => current % 2 == 0 ? current + sum : 0 + sum, 0)
return array1.map(num => num < 10 ? num + oddSum : num + oddEven)
}
console.log(addingAllTheWeirdStuff(array1, array2))

Answer 2

如果你把挑战分解成更小的部分，你可以更好地解构它，并提出你的解决方案。

这就是我所做的。。。我将很快添加更多评论和更多解释

我选择继续使用循环，因为我认为这是挑战的重点(练习循环、多个条件等(-换句话说，我选择不故意使用map / reduce，但如果允许的话，请使用@charmful0x的答案，因为这会减少代码：(

// function to get sum of all odd numbers in array
function getSumOfAllOddNumbersInArray( elementArray ){
var sumOfOddNumbers = 0;

for (let i = 0; i < elementArray.length; i++) {
// use remainder operator to find out if element is odd or not
if (elementArray[i] % 2 !== 0 ) {
sumOfOddNumbers += elementArray[i];
}
}

return sumOfOddNumbers;
}
// function to get sum of all EVEN numbers in array
function getSumOfAllEvenNumbersInArray( elementArray ){
var sumOfEvenNumbers = 0;

for (let i = 0; i < elementArray.length; i++) {
// use remainder operator to find out if element is odd or not
if (elementArray[i] % 2 === 0 ) {
sumOfEvenNumbers += elementArray[i];
}
}

return sumOfEvenNumbers;
}
// Return true if there is at least one element in array that is greater than 20
function hasElementOverTwenty( elementArray ){
for (let i = 0; i < elementArray.length; i++) {
if (elementArray[i] > 20 ) {
// no need to keep looping, we found one - exit function
return true;
}
}

return false;
}

function addingAllTheWeirdStuff( firstArray, secondArray ){
var sumOfOddNumbersInArray = getSumOfAllOddNumbersInArray( secondArray );
var sumOfEvenNumbersInArray = getSumOfAllEvenNumbersInArray( secondArray );
var needToAddOne = hasElementOverTwenty( secondArray );

for (let i = 0; i < firstArray.length; i++) {
// Challenge One
if (firstArray[i] < 10) {
firstArray[i] = firstArray[i] + sumOfOddNumbersInArray;       

} else if (firstArray[i] > 10) {
// Challenge Two
firstArray[i] = firstArray[i] + sumOfEvenNumbersInArray;       
}

// bonus
if( needToAddOne ){
firstArray[i]++;
}   
}

return firstArray;
}
// Uncomment these to check your work!
console.log(addingAllTheWeirdStuff([1, 3, 5, 17, 15], [1, 2, 3, 4, 5])); 
console.log('expected:' + [10, 12, 14, 23, 21] );
console.log(addingAllTheWeirdStuff([1, 3, 5, 17, 15, 1], [1, 2, 3, 4, 5, 22]));
console.log('expected:' + [11, 13, 15, 46, 44, 11] );

Challenge question says: Write a function addingAllTheWeirdStuff which adds the sum of all the odd numbers in array2 to each element under 10 in array1.
Similarly, addingAllTheWeirdStuff should also add the sum of all the even numbers in array2 to those elements over 10 in array1.
BONUS: If any element in array2 is greater than 20, add 1 to every element in array1.