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I am trying to use the following code with map or filter methods is it possible to do that:

function doubleOddNumbers(numbers) {
  const newNumbers = [];
  for (let i = 0; i < numbers.length; i++) {
    if (numbers[i] % 2 !== 0) {
      newNumbers.push(numbers[i] * 2);
    }
  }
  return newNumbers;
}

const myNumbers = [1, 2, 3, 4];
console.log(doubleOddNumbers(myNumbers);

// Do not change or remove anything below this line
module.exports = {
  myNumbers,
  doubleOddNumbers,
};

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1 Answer

You could filter the odd numbers and return the doubled value using map

function doubleOddNumbers(numbers) {
  return numbers.filter(n => n % 2).map(n => n * 2)
}

/* This would also work
function doubleOddNumbers(numbers) {
  return numbers.reduce((r, n) => n % 2 ? r.concat(n * 2) : r, [])
}
*/
const myNumbers = [1, 2, 3, 4];
console.log(doubleOddNumbers(myNumbers));

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
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