Sword refers to an offer---a character that appears only once in an array

topic:

In an integer array, except for two digits, the other numbers appear two times, please write out the program to find the two occurrences of only one number, requiring time complexity O (n), space complexity of O (1). parsing:

void FindNum (int* arr, int length)
{
	if (arr = = NULL | | | length < 2)
	{return
		;
	}
	int tty = arr[0];
	for (int i = 1; i < length; ++i)
	{
		tty = Tty^arr[i];
	}
	At this time out of the loop, TTY for two occurrences of only one number of the difference or value
	//Now need to determine the minimum bit of TTY 1 at which bit
	int index = 0;
	int num1 = 0;
	int num2 = 0;
	for (int j = 0; J < 32;++j)
	{
		if (tty& (1 << index)) = = (1 << index))
		{
			break;
  
   }
		Else
		{
			++index;
		}
	}
	At this point, you need to divide the two occurrences only once into two groups for
	(int p = 0; p < length; ++p)
	{
		if (arr[p]& (1 << index) = = (1 <& Lt Index))
		{
			num1 = num1^arr[p];
		}
		else
		{
			num2 = num2^arr[p];
		}
	}
	cout << num1 << Endl;
	cout << num2 << Endl;
}
  

void Test ()
{
	int arr[] = {1, 2, 3, 4, 5, 3, 2, 1};
	int length = sizeof (arr)/sizeof (arr[0));
	FindNum (arr, length);
}

int main ()
{
	Test ();
	System ("pause");
	return 0;
}