let descend = Object.assign([],array);

上面的行不会通过引用复制数组。

function minMinMax(array) {
  let minAbsent = 0;
  let descend = Object.assign([],array);
  // sort the array from smallest to largest
  let smallestToLargest = array.sort((a,b) => a - b)
  console.log(smallestToLargest)
  const smallest = smallestToLargest[0]
  // sort the array from largest to smallest
  let largestToSmallest = descend.sort((a,b) => b - a)
  const largest = largestToSmallest[0]
  // use second smallest number as starting point for loop,then after each number check 
  // to see if that number is in the array,if not then add the value to our array
  for (i = 1,j = smallestToLargest[1]; i < smallestToLargest.length; i++,j++) {
    if (j !== smallestToLargest[i]) {
      minAbsent = j
    }
    console.log(smallestToLargest[i])
    console.log('this is the min absent ' + minAbsent)
  }
  return [smallest,minAbsent,largest]
}

我认为这样会很好。

问题似乎是，尽管您有三个单独的指针“数组”，“ smallestToLargest”和“ largestToSmallest”，但它们都引用了相同的实际数组对象。 “ array.sort（...）”对调用方法的数组进行排序，并返回对该数组的引用。

如果在创建largestToSmallest之后尝试将smallestToLargest记录到控制台，您会看到smallestToLargest将以降序打印-因为在调用第二个array.sort时它将按降序排序。

let smallestToLargest = array.sort((a,b) => a - b)
console.log(smallestToLargest);
const smallest = smallestToLargest[0]
// sort the array from largest to smallest
let largestToSmallest = array.sort((a,b) => b - a)
console.log(smallestToLargest);

在随机数组上进行测试，得出：

[1,3,14,29,311,323]
[323,1]

要保留“ smallestToLargest”数组，可以使用“ slice（0）” 对其进行克隆。。

尝试一下：

function minMinMax(array) {
    let minAbsent = 0 
    // sort the array from smallest to largest
    let smallestToLargest = array.sort((a,b) => a - b).slice(0);
    console.log(smallestToLargest);
    const smallest = smallestToLargest[0]
    // sort the array from largest to smallest
    let largestToSmallest = array.sort((a,b) => b - a).slice(0);
    console.log(smallestToLargest);
    const largest = largestToSmallest[0]
    // use second smallest number as starting point for loop,then after each number check 
    // to see if that number is in the array,if not then add the value to our array
    for (i = 1,j = smallestToLargest[i]) {
      if (j !== smallestToLargest[i]) {
        minAbsent = j
        }
        console.log(smallestToLargest[i])
        console.log('this is the min absent ' + minAbsent)
      }
  console.log(smallestToLargest);
    return [smallest,largest]
  };