不断分段找中点，判断中点是否是要寻找的num

// 获取 num 在 vector 中的 index
int findIndex(const vector<int>& vector, int num);

int main(int argc, char *argv[])
{
    vector<int> v;
    for(int i=0; i<9; i++)
    {
        v.emplace_back();
        auto& num = v.back();
        num = i*2;
        cout << "  v.back: "<< v.back();
    }
    int id = findIndex(v, 6) ;
    if(id < 0 )
        cout << "no such number !" <<endl;
    else
        cout << "index of number is : "<< id <<endl;
    return 0;
}

int findIndex(const vector<int>& vector, int num)
{
    Q_ASSERT(vector.size() >0 );
    int size = vector.size();
    if(vector.at(0) ==num )    return 0;
    if(size==1 && vector.at(0) !=num )    return -1;
    if(num == vector.back() )   return size-1;

    int begin, mid, end;
    begin =0; end = vector.size()-1;
    mid = (begin + end) / 2;
    int count = 0;
    while(num != vector.at(mid) )
    {
        count++;
        if(count > log2(size) )
        {
            return -1;
            break;
        }
        // 比较mid和num, 按情况只取一段进行递归比较，这样才降低复杂度
        if( num > vector.at(mid) )
        {
            begin = mid; end = end;
            mid = (begin + end) / 2;
        }
        else
        {
            begin = begin; end = mid;
            mid = (begin + end) / 2;
        }
    }
    return mid;
}