剑指40-数组中只出现一次的数字

class Solution {
public:
    void FindNumsAppearOnce(vector<int> data, int* num1, int *num2) {
        map<int, int> map;
        for (int i = 0; i < data.size(); i++)
            map[data[i]]++;
        vector<int>v;
        for (int i = 0; i < data.size(); i++)
            if (map[data[i]] == 1)
                v.push_back(data[i]);
        *num1 = v[0]; *num2 = v[1];
    }
};

class Solution2 {
public:
    void FindNumsAppearOnce(vector<int> data, int* num1, int *num2) {
        int ret = 0;
        for (const int k : data) ret ^= k;
        ret &= (-ret); //找ret二进制中为1的位置
        *num1 = 0, *num2 = 0;
        for (const int k : data) { //再分别异或一次
            if (k & ret) *num1 ^= k; 
            else *num2 ^= k;
        }
    }
};