def test_single_number2(self):
self.assertEqual(3, single_number2([4, 2, 3, 2, 1, 1, 4, 2, 4, 1]))
single = random.randint(1, 100000)
nums = [random.randint(1, 100000) for _ in range(1000)]
nums *= 3 # nums contains triplets of random integers
nums.append(single)
random.shuffle(nums)
self.assertEqual(single, single_number2(nums))
def test_single_number2(self):
self.assertEqual(3, single_number2([4, 2, 3, 2, 1, 1, 4, 2, 4, 1]))
single = random.randint(1, 100000)
nums = [random.randint(1, 100000) for _ in range(1000)]
nums *= 3 # nums contains triplets of random integers
nums.append(single)
random.shuffle(nums)
self.assertEqual(single, single_number2(nums))
""" Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. Limitation: Time Complexity: O(N) and Space Complexity O(1) For example: Given nums = [1, 2, 1, 3, 2, 5], return [3, 5]. Note: The order of the result is not important. So in the above example, [5, 3] is also correct. Solution: 1. Use XOR to cancel out the pairs and isolate A^B 2. It is guaranteed that at least 1 bit exists in A^B since A and B are different numbers. ex) 010 ^ 111 = 101 3. Single out one bit R (right most bit in this solution) to use it as a pivot 4. Divide all numbers into two groups. One group with a bit in the position R One group without a bit in the position R 5. Use the same strategy we used in step 1 to isolate A and B from each group. """ # Two element once from algorithms.bit import single_number,single_number2,single_number3 nums = [1, 2, 1, 3, 2, 5] print(single_number(nums)) ## XXX print(single_number2(nums)) ## XXX print(single_number3(nums))
""" Given an array of integers, every element appears three times except for one, which appears exactly once. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? Solution: 32 bits for each integer. Consider 1 bit in it, the sum of each integer's corresponding bit (except for the single number) should be 0 if mod by 3. Hence, we sum the bits of all integers and mod by 3, the remaining should be the exact bit of the single number. In this way, you get the 32 bits of the single number. """ ## one once from algorithms.bit import single_number2 a = [4, 1, 2, 1, 2] b = [2, 2, 1] print(single_number2(a)) print(single_number2(b))