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?

public int singleNumber(int[] nums) {
int result = 0;
for (int i = 0; i < nums.length; i++) {
result ^= nums[i];
}
return result;
}

注意到A^A=0，所以A^B^A=B