Given an 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?



In [7]:

    
class Solution(object):
    def singleNumber(self, nums):
        """
        通过相同数字异或为0这个知识点进行操作
        和0异或等于本身来初始化ans
        :type nums: List[int]
        :rtype: int
        """
        ans = 0
        for n in nums:
            ans ^= n
        return ans



In [8]:

    
Solution().singleNumber([2, 4, 5 , 5, 4])









    Out[8]:





2



In [9]:

    
Solution().singleNumber([2]) # answer is 2









    Out[9]:





2



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