谢谢你这个有趣的问题:

因为你让我想起了牛顿的工作，它真的可以解决这个问题

请参考牛顿恒等式

As变量的数量=方程的数量(必须为一致性)

我认为，对于这个问题，我们应该提高袋数的幂，以便创建不同的方程。

我不知道，但是，我相信如果有一个函数，比如f，我们要加上f(xi)

x1+x2+…+ xk = z1

x12 + x22 + ... + xk2 = z2

............

............

............

x1k + x2k + ... + xkk = XP

休息是一个不确定时间和空间复杂性的数学工作，但牛顿恒等式肯定会发挥重要作用。

我们不能用集合理论吗 .difference_update()或在这个问题方法中是否有线性代数的机会?

你可以通过阅读Muthukrishnan的几页-数据流算法:谜题1:寻找缺失的数字来找到它。它准确地显示了您正在寻找的泛化。也许这就是面试官读到的内容，也是他提出这些问题的原因。

还请参阅sdcvvc的直接相关答案，其中还包括伪代码(万岁!没有必要阅读那些棘手的数学公式:)(谢谢，干得好!)

    //sort
    int missingNum[2];//missing 2 numbers- can be applied to more than 2
    int j = 0;    
    for(int i = 0; i < length - 1; i++){
        if(arr[i+1] - arr[i] > 1 ) {
            missingNum[j] = arr[i] + 1;
            j++;
        }
    }

我不知道这是否有效，但我想建议这个解决方案。

计算这100个元素的xor 计算98个元素的xor(在2个元素被移除之后) 现在(1的结果)XOR(2的结果)给你两个缺失的no的XOR，如果a和b是缺失的元素 4.用常用的求和公式diff得到缺失的no的和，我们设diff是d。

现在运行一个循环，得到可能的对(p,q)，它们都位于[1,100]，和为d。

当获得一对时，检查(3的结果)是否XOR p = q 如果是，我们就完成了。

如果我错了，请纠正我，如果这是正确的，也请评论时间复杂性

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。

也许这个算法可以解决问题1:

预计算前100个整数的xor (val=1^2^3^4....100) 对来自输入流的元素进行Xor (val1=val1^next_input) 最终的答案= val ^ val1

或者更好:

def GetValue(A)
  val=0
  for i=1 to 100
    do
      val=val^i
    done
  for value in A:
    do
      val=val^value 
    done
  return val

这个算法实际上可以扩展到两个缺失的数字。第一步还是一样。当我们调用缺少两个数字的GetValue时，结果将是a1^a2是缺少的两个数字。让说

跌倒 = a1^a2

Now to sieve out a1 and a2 from val we take any set bit in val. Lets say the ith bit is set in val. That means that a1 and a2 have different parity at ith bit position. Now we do another iteration on the original array and keep two xor values. One for the numbers which have the ith bit set and other which doesn't have the ith bit set. We now have two buckets of numbers, and its guranteed that a1 and a2 will lie in different buckets. Now repeat the same what we did for finding one missing element on each of the bucket.