problem：
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?
thinking：
（1）考察位运算，C/C++的异或运算符为 ^
0^a=a;
a^a=0;
a^b=b^a;
(2)这道题的解法就出来了：n个数的异或结果就是待求数
code：
class Solution {
public:
int singleNumber(vector<int>& nums) {
int n=nums.size();
int ret=nums[0];
for(int i=1;i<n;i++)
ret^=nums[i];
return ret;
}
};
原文地址：http://blog.csdn.net/hustyangju/article/details/45367881
