Array     C++     Design     Easy     Hash Function     Hash Table     Linked List    

Problem Statement:

Design a HashMap without using any built-in hash table libraries.

Implement the MyHashMap class:

  • MyHashMap() initializes the object with an empty map.
  • void put(int key, int value) inserts a (key, value) pair into the HashMap. If the key already exists in the map, update the corresponding value.
  • int get(int key) returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
  • void remove(key) removes the key and its corresponding value if the map contains the mapping for the key.

 

Example 1:

Input
["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
Output
[null, null, null, 1, -1, null, 1, null, -1]

Explanation
MyHashMap myHashMap = new MyHashMap();
myHashMap.put(1, 1); // The map is now [[1,1]]
myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
myHashMap.get(1);    // return 1, The map is now [[1,1], [2,2]]
myHashMap.get(3);    // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
myHashMap.get(2);    // return 1, The map is now [[1,1], [2,1]]
myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
myHashMap.get(2);    // return -1 (i.e., not found), The map is now [[1,1]]

 

Constraints:

  • 0 <= key, value <= 106
  • At most 104 calls will be made to put, get, and remove.

Solution:

We learn this in college. It is good to get a chance to implement. Here we use modulo operator as the hashing function and M is chosen to be a big enough prime.

 
class Node {
    public:
    int key;
    int val;
    Node *next;
    Node(int k, int v, Node *n){key=k;val=v;next=n;}
};

class MyHashMap {
public:
    vector<Node *>nodes;
    int N;
    MyHashMap(int n=99991) {
        nodes = vector<Node *>(n,NULL);
        N = n;        
    }
    
    void put(int key, int value) {
        int h = key%N;
        if (nodes[h]==NULL)
        {
            nodes[h]=new Node(key,value,NULL);
            return;
        }
        Node *curr=nodes[h];
        while (curr!=NULL)
        {
            if (curr->key==key)
            {
                curr->val=value;
                return;
            }
            curr=curr->next;
        }
        while (curr->next!=NULL) curr=curr->next;
        curr->next=new Node(key,value,NULL);
    }
    
    int get(int key) {
        int h = key%N;
        if (nodes[h]==NULL) return -1;
        Node *curr = nodes[h];
        while (curr!=NULL)
        {
            if (curr->key==key) return curr->val;
            curr=curr->next;
        }
        return -2;
    }
    
    void remove(int key) {
        int h=key%N;
        if (nodes[h]==NULL) return;
        if (nodes[h]->key==key) 
        {
            nodes[h]=nodes[h]->next;
            return;
        }
        Node *curr=nodes[h];
        while (curr->next!=NULL && curr->next->key!=key) curr=curr->next;
        curr->next = curr->next->next;        
    }
};

/**
 * Your MyHashMap object will be instantiated and called as such:
 * MyHashMap* obj = new MyHashMap();
 * obj->put(key,value);
 * int param_2 = obj->get(key);
 * obj->remove(key);
 */

I solved the HashSet problem too using similar method: https://leetcode.com/problems/design-hashset/discuss/1968599/Hashing-using-modulo