Leetcode: Single Number

Single Number | LeetCode OJ

Since every number is repeating twice we can use xor to eliminate the duplicates as n xor n = 0.

So after taking xor of all numbers only the number that appears once will remain

O(1) space, O(n) time complexity

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res = nums[0]
        for n in nums[1:]:
            res = res ^ n
        return res