LeetCode: Single Number

Q: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?

思路:一个整数异或它本身,结果为0. 所以只要遍历一次数组即可得到只出现一次的数字.

代码:

	public int singleNumber(int[] nums) {
        int length = nums.length, i = 0;
		long sum = 0L;
		while (i < length) {
			sum ^= nums[i++];
		}
		return (int) sum;
    }