[C++]LeetCode: 66 Single Number
题目：
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：
要求线性时间完成，并且不使用额外的存储空间。
**非常之巧妙**
使用异或来排除double值，剩下的就是single one. 异或具有可交换性，顺序可变。
比如给出数组{2,1,4,5,2,4,1} ，求异或，2^1^4^5^2^4^1 <=> (2^2)^(1^1)^(4^4)^(5) <=> (0)^(0)^(0)^5 = 5
AC Code:
class Solution {
public:
int singleNumber(int A[], int n) {
int ret;
ret = A[0];
for(int i = 1; i < n; i++)
{
ret ^= A[i];
}
return ret;
}
};
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