Finding Missing Number in C#

static int findmissingnum(int[] num) //1 2 3 5
{
if (num == null) throw new ArgumentNullException("num");
if (num.Length == 0) throw new ArgumentOutOfRangeException("num");

if (num[num.Length - 1] == num.Length)
return 0;

int startIndex = 0;
int endIndex = num.Length - 1;
int midNumber = num.Length / 2;

while (midNumber >= startIndex && midNumber <= endIndex)
{
if (num[midNumber] == midNumber + 1)
{
if (startIndex == midNumber)
{
break;
}
startIndex = midNumber;
midNumber = (midNumber + num.Length - 1) / 2;
}
else
{
if (startIndex == midNumber)
{
midNumber--; break;
}

endIndex = midNumber;
midNumber = (midNumber + startIndex) / 2;
}
}
if (midNumber == -1)
return 1;

return num[midNumber] + 1;
}

Test Cases:
1. null
2. zero length int []
3. first numer is missing 2,3,4,5
1,2,4,5
1,3,4,5
1,2,3,5
1 and 1,2 and 1,2,3 and 1,2,3,4and 1,2,3,4,5 and 1,2,3,4,5,6

1,2,4,5
1,3,4,5
1,2,3,5
1 and 1,2 and 1,2,3 and 1,2,3,4and 1,2,3,4,5 and 1,2,3,4,5,6
if (num[num.Length - 1] == num.Length)
return 0;
1 1,2 1,2,3 1,2,3,4
thead safe
int[] num
1M
int32.MaxValue
2^31
2^31-1
stable
environment
doc