使用Javascript查找数组中最小的未使用数字

Question

I have an array full of numbers. Here's an example:

myArray = [0,1,2,4,5];

I need to find the lowest unused number starting from 1, so in this case it will be 3.

I've been reading up of using indexOf but I'm unsure how to use it for my specific purpose.

Answer 1

Assuming the array isn't sorted, you always start at 0, and taking into account your desire to find a highest number if there isn't one missing:

var k = [6, 0, 1, 2, 4, 5];

k.sort(function(a, b) { return a-b; });   // To sort by numeric

var lowest = -1;
for (i = 0;  i < k.length;  ++i) {
  if (k[i] != i) {
    lowest = i;
    break;
  }
}
if (lowest == -1) {
    lowest = k[k.length - 1] + 1;
}
console.log("Lowest = " + lowest);

Logs answer 3. If 3 was also in there, would log 7 since no other number is missing.

If you aren't always starting at zero, use an offset:

var k = [6, 2, 3, 4, 5];

k.sort(function(a, b) { return a-b; });   // To sort by numeric

var offset = k[0];
var lowest = -1;
for (i = 0;  i < k.length;  ++i) {
  if (k[i] != offset) {
    lowest = offset;
    break;
  }
  ++offset;
}
if (lowest == -1) {
    lowest = k[k.length - 1] + 1;
}
console.log("Lowest = " + lowest);

Logs answer 7 since none are missing after 2 which starts the sequence.

Answer 2

This takes a sequence starting from a number (like 1 in your example) and returns the lowest unused number in the sequence.

function lowestUnusedNumber(sequence, startingFrom) {
  const arr = sequence.slice(0);
  arr.sort((a, b) => a - b);

  return arr.reduce((lowest, num, i) => {
    const seqIndex = i + startingFrom;
    return num !== seqIndex && seqIndex < lowest ? seqIndex : lowest
  }, arr.length + startingFrom);
}

Example:

> lowestUnusedNumber([], 1)
1
> lowestUnusedNumber([1,2,4], 1)
3
> lowestUnusedNumber([3], 1)
1

In exchange for readability, it's slightly less optimized than the other example because it loops over all items in the array instead of breaking as soon as it finds the missing item.