136. Single Number-位运算，异或

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?

这题用到位运算异或

异或有下面的性质：

1、交换律：a ⊕ b = b ⊕ a

2、结合律:   a ⊕b ⊕ c = a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c;

3、对于任何数x，都有x^x=0，x^0=x

4、自反性: A XOR B XOR B = A xor  0 = A

假设nums = [2,1,3,1,2]

那么res = 0^2^1^3^1^2 = 0^(1^1)^(2^2)^3 = 3 

代码：

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int result = 0, n = nums.size();
        for (int i = 0; i<n; i++)
    		result ^= nums[i];     //利用异或的交换律：a^b = b^a 以及 结合律：a^b^c=(a^b)^c=a^(b^c)
    	return result;
    }
};