如何提升奇数并将偶数保持在原位

Question

As you can see in my code, the odd numbers are ascended, but the even numbers are eliminated. I want the even numbers to stay in their places. The expected log would be [1, 3, 2, 8, 5, 4]

 function sortArray(array) { let sortedNumbers = array.sort(); let newArray = []; for (let i = 0; i < sortedNumbers.length; i++) { if (sortedNumbers[i] % 2 !== 0) { newArray.push(sortedNumbers[i]); } } return newArray; } console.log(sortArray([5, 3, 2, 8, 1, 4]));

Answer 1

I think the clearest way to do this would be to extract all the odd numbers into a separate array, sort that array, then insert them back into the original array:

 function sortArray(array) { const odds = array.filter(num => num % 2 === 1); odds.sort((a, b) => a - b); return array.map( num => num % 2 === 1 ? odds.shift() : num ); } console.log(sortArray([5, 3, 2, 8, 1, 4]))

Note that you can't use .sort , because .sort compares lexicographically (eg 11 will come before 2 , which is wrong) - use .sort((a, b) => a - b); instead.

Answer 2

I'm not sure what "ascended" means, but for instance if you wanted to add 0.1 to each odd number while leaving the evens as they are, you could simply add an else statement as follows.

Note that providing the numericSort function is necessary unless you want your numbers to be sorted alphabetically (which is the default for the .sort method.)

 const numericSort = (a, b) => a - b; function sortArray(arr) { const sortedNumbers = arr.sort(numericSort), newArray = []; for(num of sortedNumbers) { if (num % 2 !== 0) { newArray.push(num + 0.1); } else { newArray.push(num); } } return newArray; } console.log(sortArray([5, 31, 20, 8, 1, 4]))