Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Example：
Input: [2,2,1]
Output: 1
Input: [4,1,2,1,2]
Output: 4
Note：
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
解释下题目：
给定一个非空数组，其中每个数组都是有两个的，只有一个数字只有一个，找出那个数字
1. 抑或操作
实际耗时：0ms
public int singleNumber(int[] nums) {
//异或操作会留下最后的那一个数
int result = nums[0];
for (int i = 1; i < nums.length; i++) {
result ^= nums[i];
}
return result;
}
两个相同的数字异或是0
