Single Number

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?

解法：

相同的两个数 按位异或 结果为0,

则 可以把所有数字异或后的结果 为单独的数字。

 int singleNumber(vector<int>& nums) {
        int size=nums.size();
        if(size==0)
            cout<<"error";
        int result=nums[0];
        for(int i=1;i<size;++i){
            result^=nums[i];
        }
        return result;
    }

Single Number

原文：http://searchcoding.blog.51cto.com/1335412/1751208