Suppose x = 0 and your list is [1,2,3,2,1]. If you xor 'x' with every number in your list then x = 1^2^3^2^1, which is the same as x = (1^1)^(2^2)^3 = 0^0^3 = 3. So, x=3 is the number that wasn't repeated. This works because a^b = b^a and a^a = 0.

Iterate through the list bitwise-XORing the numbers together into an accumulator. Because x ^ x = 0, and x ^ 0 = x, and ^ is commutative (x ^ y = y ^ x), the final value in the accumulator for a list like [x, y, y, x, z] is:

    x ^ y ^ y ^ x ^ z 
  = x ^ x ^ y ^ y ^ z
  = 0 ^ 0 ^ z
  = z

Here's an example in python:

  >>> import random, operator
  >>> nums = (range(5)*2)
  >>> random.shuffle(nums)
  >>> nums.pop() 
  1
  >>> nums
  [0, 4, 0, 3, 2, 4, 3, 1, 2]
  >>> reduce(operator.xor, nums)
  1