您可以在O（n）的做到这一点。遍历数组和计算所有数字的总和。现在，自然数之和从1到N，可以pssed为 EX $ P $ NX（N + 1）/ 2 。在你的情况下，N = 100。

You can do this in O(n). Iterate through the array and compute the sum of all numbers. Now, sum of natural numbers from 1 to N, can be expressed as Nx(N+1)/2. In your case N=100.

从减去阵列的总和NX（N + 1）/ 2 ，其中N = 100。

Subtract the sum of the array from Nx(N+1)/2, where N=100.

这是缺少的数。可以在其中的总和计算的迭代过程中被检测到的空槽。

That is the missing number. The empty slot can be detected during the iteration in which the sum is computed.

// will be the sum of the numbers in the array.
int sum = 0;
int idx = -1;
for (int i = 0; i < arr.length; i++)
{
    if (arr[i] == 0)
    {
         idx = i; 
    }
    else 
    {
         sum += arr[i];
    }
}

// the total sum of numbers between 1 and arr.length.
int total = (arr.length + 1) * arr.length / 2;

System.out.println("missing number is: " + (total - sum) + " at index " + idx);