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?
int singleNumber(int A[], int n) {
int x;
for(size_t i=0;i<n;++i)
{
x^=A[i];//异或两次相同的值为0,异或0为原值
}
return x;
}