15算法课程 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?

solution:

数组中只有一个数出现一次，其他的数都出现两次，异或后的结果就是那个单次出现的整数.

code:

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