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?

class Solution {

public:

    int singleNumber(vector<int>& nums) {

        for(int i=1;i<nums.size();i++){

            nums[0]^=nums[i];

        }

        return nums[0];

    }

};

要求O（n）并且不用额外的变量。
非常巧妙的技巧。相同数取异或为0，所0和所有数的异或为本身，所以最后剩下的就是Single Number。

136. Single Number leetcode做题报告