136. 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?

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class Solution {public:    int singleNumber(vector<int>& nums) {       for (int  i = 1; i !=nums.size(); i++)    {        nums[0] ^= nums[i];    }    return nums[0];    }};

解释：当使用异或，无论本来是1 还是 0 ，当遇到偶数个1或者偶数个0，那么那些偶数个1或者偶数个0都会自己对消，无论这些偶数个1,0使用异或操作的先后顺序如何。