Single Number leetcode

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?

1. a ^ a = 0

2. a ^ b = b ^ a

3. a ^ b ^ c = a ^ (b ^ c) = (a ^ b) ^ c

----> a ^ b ^ a = b

所以把全部数字异或一遍就能找到只出现一次的数 时间O(n)

public class Solution {
    public int singleNumber(int[] nums) {
        if (nums.length == 0 || nums == null) {
            return -1;
        }
        int res = 0;
        for (int i = 0; i < nums.length; i++) {
            res = res ^ nums[i];
        }
        return res;
    }
}