Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1] Output: 1
Example 2:
Input: [4,1,2,1,2] Output: 4
# Python3
class Solution:
def singleNumber(self, nums: List[int]) -> int:
#s = set()
#for n in nums:
# if n in s:
# s.remove(n)
# else:
# s.add(n)
#for n in s:
# return n
#return None
#nums.sort()
#index = 0
#while index + 1 < len(nums):
# if nums[index] != nums[index+1]:
# return nums[index]
# index += 2
#return nums[-1]
a = 0
for i in nums:
a ^= i
return a