Single Number leetcode

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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这个题我首先想到的是用hashmap，第一次添加，第二次删除，剩下的没有删除的就是答案

int singleNumber(vector<int>& nums) {
    unordered_map<int, int> cnt;
    for (auto i : nums)
    {
        if (cnt.find(i) != cnt.end())
            cnt.erase(i);
        else
            cnt.insert(make_pair(i, ));
    }
    return (*cnt.begin()).first;
}

去discuss中，发现还有更简单的解法，使用XOR位运算，因为A XOR A = 0，而且XOR具有结合性，所以A XOR B XOR C XOR A XOR C XOR B = 0

int singleNumber(vector<int>& nums) {
    int result = ;
    for (int i = ; i < nums.size(); i++)
    {
        result ^= nums[i];
    }
    return result;
}