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?
解法：
相同的两个数 按位异或 结果为0,
则 可以把所有数字异或后的结果 为单独的数字。
int singleNumber(vector<int>& nums) {
int size=nums.size();
if(size==0)
cout<<"error";
int result=nums[0];
for(int i=1;i<size;++i){
result^=nums[i];
}
return result;
}
