C ++中的哈希表实现-如何返回具有特定值的键

Question

I am trying to solve a programming question using Hash Tables in C++. It is supposed to be fairly simple implementation of hash tables. I am extremely new to Hash Tables and this is my first try at implementation.

The question is that I have been given an array which contains integers. All but one integer repeats itself twice. I have to find an integer that doesn't.

Input: {1,2,1,3,3}
Output: 2

My solution is that I will start putting these keys inside a hash table and if I find a key inside the hash table beforehand, I will delete that key from the hash table.

My code implementation works but I now I wanted to see how I can get back the right value (2 in this case) since even after erasing the key, the keys remain with value 0.

Here is my code:

int main()
{

    int num[5] = {1,2,1,3,3};

    map <int,int> mymap;

    for(int i=0;i<5;i++)
    {
      if(mymap.find(num[i])!=mymap.end())
        {  
          mymap.erase(num[i]);
        }
      else
        {
          mymap[num[i]] = 10; //10 is just a placeholder value. 
        }
    }
    cout<<"0:"<<mymap[0]<<endl;
    cout<<"1:"<<mymap[1]<<endl;
    cout<<"2:"<<mymap[2]<<endl;   //Should only find 10 value in 2
    cout<<"3:"<<mymap[3]<<endl;
    cout<<"4:"<<mymap[4]<<endl;
}

Output:

0:0
1:0
2:10
3:0
4:0

Answer 1

std::map is a tree, not a hash table. For a hash table you want std::unordered_map .

But to answer your question:

You can use the map iterator to get the only remaining value:

if (!mymap.empty()) {
    cout << mymap.begin()->first;
}

But beware - when you call cout << mymap[X] , it also adds X to the map. So remove all those debugging lines at the end.

And by the way - when you don't have a value, just a key, then you can use a set instead (or unordered_set ).

Answer 2

Just increment the value in the map, as the integers are default constructed (and thus initialized to 0 ):

int num[5] = {1,2,1,3,3};

unordered_map <int,int> mymap;

for(int i=0;i<5;i++)
{
   ++mymap[num[i]];
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl;   //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;