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];//异或两次同样的值为0,异或0为原值 } return x; }