leetcode136

题目

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

解答:

因为只有一个数字是出现了一次的,其他的都是出现了两次,那么可以用按位异或,因为任意一个数字和其自己按位异或得到的结果都是0.

所以最后的那个数就是结果

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