c++ - 如何在数组中找到两个缺失值？

Question

可能重复：
简单的​​面试问题变得更难：给定数字 1..100，找到缺失的数字

一个求职面试题。假设我们有一个大小为 N-2 的数组，所有值都从 1 到 N，除了两个缺失值。(N>0)

需要一种用于查找两个缺失数字的算法，该算法仅遍历数组一次。

Answer 1

// Receiving array with values from 1 to N (NOT 0 to N-1)
// N is max value & # of values, NOT size of arr (arr is in size N-2)
void Find2Numbers(unsigned arr[], size_t N, unsigned & n1, unsigned & n2)
{
    unsigned sum = N*(N+1)/2;  // sum of elements
    double pro = 1;  // Products will be calculated within the loop, because fact(N) can be very large

    for(size_t i = 0 ; i < N-2 ; i++)
    {
        pro *= (i+1);  // mult by i+1 to get factorial
        pro /= arr[i];  // divide by arr[i] to find n1*n2
        sum -= arr[i];
    }
    pro *= (N-1)*N;  // 2 missing indexes, as arr is missing 2 elements
    // sum = n1+n2
    // pro = n1*n2  =>
    n1 = (sum+sqrt(sum*sum-4*pro))/2;
    n2 = (sum-sqrt(sum*sum-4*pro))/2;
}