Leetcode:single-number

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?

由于要求时间复杂度为线性，故采用按位异或求法。

思路：

For anyone who didn't understood why this works here is an explanation. This XOR operation works because it's like XORing all the numbers by itself. So if the array is {2,1,4,5,2,4,1} then it will be like we are performing this operation

((2^2)^(1^1)^(4^4)^(5)) => (0^0^0^5) => 5.

Hence picking the odd one out ( 5 in this case).(来源于Leetcode Discuss)

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?