Bits: the single number

Every value in the list appears exactly twice, except one value that appears once. Find that lonely value.

You could count occurrences with a dictionary, but there's a slicker trick using the XOR operator ^. XOR has two magic properties:

x ^ x == 0     # anything XOR'd with itself cancels to 0
x ^ 0 == x     # XOR with 0 leaves a value unchanged

XOR is also commutative and associative, so the order doesn't matter. If you XOR the whole list together, every pair cancels to 0, and the single value is all that survives:

2 ^ 2 ^ 1  ==  (2 ^ 2) ^ 1  ==  0 ^ 1  ==  1

This runs in one pass with no extra memory.

Your task

Write single_number(nums) returning the one value that appears only once.

single_number([2, 2, 1])        # 1
single_number([4, 1, 2, 1, 2])  # 4
single_number([7])              # 7

Hint: start an accumulator at 0 and XOR every element into it.