一种方法是对质数101取模。

计算并存储整数1到100的乘积，对该数字取101的模。小外显:结果是1。

计算并存储所有数字1到100的和，对结果对101进行模运算。小exo:结果是0。

现在假设袋子里的数字x和y都被拿走了。

求包里所有东西对101模的乘积和。这样我就知道的值

A = x+y和 b = x * y

modulo 101。

现在很容易求出x和y对101的模(求解含有101个元素的有限域上的二次多项式)。

现在你知道了x和y对101的模。但是因为你知道x和y都小于101，所以你知道它们的真实值。

一个非常简单的Q2解决方案，我很惊讶没有人回答。用Q1的方法求两个缺失数字的和。我们用S表示它，那么缺失的数字中一个比S/2小另一个比S/2大(胡说)将从1到S/2的所有数字相加，并将其与公式的结果进行比较(类似于Q1中的方法)，以找到缺失数字之间的较低者。用S减去它，找出缺失的更大的数。

我会用另一种方法来回答这个问题，询问面试官关于他试图解决的更大问题的更多细节。根据问题和围绕它的需求，显而易见的基于集的解决方案可能是正确的，而生成一个列表然后从中挑选的方法可能不是。

For example, it might be that the interviewer is going to dispatch n messages and needs to know the k that didn't result in a reply and needs to know it in as little wall clock time as possible after the n-kth reply arrives. Let's also say that the message channel's nature is such that even running at full bore, there's enough time to do some processing between messages without having any impact on how long it takes to produce the end result after the last reply arrives. That time can be put to use inserting some identifying facet of each sent message into a set and deleting it as each corresponding reply arrives. Once the last reply has arrived, the only thing to be done is to remove its identifier from the set, which in typical implementations takes O(log k+1). After that, the set contains the list of k missing elements and there's no additional processing to be done.

这当然不是批处理预先生成的数字袋的最快方法，因为整个过程运行O((log 1 + log 2 +…)+ log n) + (log n + log n-1 +…+ log k))。但它确实适用于任何k值(即使它事先不知道)，在上面的例子中，它的应用方式使最关键的区间最小化。

我相信我有一个O(k)时间和O(log(k)空间算法，前提是你有任意大整数的下限(x)和log2(x)函数:

你有一个k位的长整数(因此是log8(k)空间)，其中你加上x^2，其中x是你在袋子里找到的下一个数字:s=1^2+2^2+…这需要O(N)时间(这对面试官来说不是问题)。最后得到j= (log2(s))这是你要找的最大的数。然后s=s-j，重复上面的步骤:

for (i = 0 ; i < k ; i++)
{
  j = floor(log2(s));
  missing[i] = j;
  s -= j;
}

现在，对于2756位的整数，通常没有floor和log2函数，而是用于double。所以呢?简单地说，对于每2个字节(或1、3、4)，您可以使用这些函数来获得所需的数字，但这增加了O(N)因素的时间复杂度

免责声明:我已经读了这个问题好几天了，但我的知识超出了我对数学的理解。

我试着用set来解决它:

arr=[1,2,4,5,7,8,10] # missing 3,6,9
NMissing=3
arr_origin = list(range(1,arr[-1]+1))

for i in range(NMissing):
      arr.append(arr[-1]) ##### assuming you do not delete the last one

arr=set(arr)
arr_origin=set(arr_origin)
missing=arr_origin-arr # 3 6 9

要解决缺少2(和3)个数字的问题，您可以修改quickselect，它平均在O(n)内运行，如果分区是就地完成的，则使用恒定内存。

Partition the set with respect to a random pivot p into partitions l, which contain numbers smaller than the pivot, and r, which contain numbers greater than the pivot. Determine which partitions the 2 missing numbers are in by comparing the pivot value to the size of each partition (p - 1 - count(l) = count of missing numbers in l and n - count(r) - p = count of missing numbers in r) a) If each partition is missing one number, then use the difference of sums approach to find each missing number. (1 + 2 + ... + (p-1)) - sum(l) = missing #1 and ((p+1) + (p+2) ... + n) - sum(r) = missing #2 b) If one partition is missing both numbers and the partition is empty, then the missing numbers are either (p-1,p-2) or (p+1,p+2) depending on which partition is missing the numbers. If one partition is missing 2 numbers but is not empty, then recurse onto that partiton.

由于只缺少2个数字，该算法总是丢弃至少一个分区，因此保持了O(n)个快速选择的平均时间复杂度。类似地，当缺少3个数字时，该算法也会在每次传递中丢弃至少一个分区(因为当缺少2个数字时，最多只有1个分区包含多个缺少的数字)。然而，我不确定当添加更多缺失的数字时，性能会下降多少。

下面是一个不使用就地分区的实现，所以这个例子不满足空间要求，但它确实说明了算法的步骤:

<?php

  $list = range(1,100);
  unset($list[3]);
  unset($list[31]);

  findMissing($list,1,100);

  function findMissing($list, $min, $max) {
    if(empty($list)) {
      print_r(range($min, $max));
      return;
    }

    $l = $r = [];
    $pivot = array_pop($list);

    foreach($list as $number) {
      if($number < $pivot) {
        $l[] = $number;
      }
      else {
        $r[] = $number;
      }
    }

    if(count($l) == $pivot - $min - 1) {
      // only 1 missing number use difference of sums
      print array_sum(range($min, $pivot-1)) - array_sum($l) . "\n";
    }
    else if(count($l) < $pivot - $min) {
      // more than 1 missing number, recurse
      findMissing($l, $min, $pivot-1);
    }

    if(count($r) == $max - $pivot - 1) {
      // only 1 missing number use difference of sums
      print array_sum(range($pivot + 1, $max)) - array_sum($r) . "\n";
    } else if(count($r) < $max - $pivot) {
      // mroe than 1 missing number recurse
      findMissing($r, $pivot+1, $max);
    }
  }

Demo