力扣算法——136SingleNumber【E】

Given a non-empty 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?

Example 1:

Input: [2,2,1]
Output: 1

Example 2:

Input: [4,1,2,1,2]
Output: 4


Solution:
　　使用异或运算，相同为0，不同为1，剩下的那个就是唯一的那个数了

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int res = 0;
        for (auto a : nums)res = res ^ a;
        return res;
    }
};