146. LRU Cache

1. Description

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.
Implement the LRUCache class:
LRUCache(int capacity) Initialize the LRU cache with positive size capacity.
int get(int key) Return the value of the key if the key exists, otherwise return -1.
void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

2. Follow Up

Could you do get and put in O(1) time complexity?

3. Example

Example 1:
Input
[“LRUCache”, “put”, “put”, “get”, “put”, “get”, “put”, “get”, “get”, “get”]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, null, -1, 3, 4]

Explanation
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // cache is {1=1}
lRUCache.put(2, 2); // cache is {1=1, 2=2}
lRUCache.get(1); // return 1
lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2); // returns -1 (not found)
lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1); // return -1 (not found)
lRUCache.get(3); // return 3
lRUCache.get(4); // return 4

4. Constraints

1 <= capacity <= 3000
0 <= key <= 3000
0 <= value <= $10^4$
At most $3 * 10^4$ calls will be made to get and put

5. Solutions

My Accepted Solution(Follow Up)

Get Time complexity: O(1)
Set Time complexity: O(1)

if we want to get or set a value with O(1) time complexity, we must use a map
at the same time, the process of delete may result in moving a node, also with O(1) time complexity, so it is a list

struct Node
{
    struct Node *prev, *next;
    int key, value;

    Node(int _key = 0, int _value = 0) : key(_key), value(_value) {prev = next = nullptr;}
};

class LRUCache 
{
private:
    unordered_map<int, Node*> nodes;
    Node *head, *tail;
    int size, capacity;

    void removeNode(Node *m_head)
    {
        m_head->prev->next = m_head->next;
        m_head->next->prev = m_head->prev;
    }

    void addToHead(Node *m_head)
    {
        m_head->next = head->next;
        m_head->prev = head;
        head->next->prev = m_head;
        head->next = m_head;
    }

    void moveToHead(Node *m_head)
    {
        removeNode(m_head);
        
        addToHead(m_head);
    }

    Node *removeLastNode()
    {
        auto node = tail->prev;
        removeNode(node);

        return node;
    }
public:
    // LRUCache(int capacity)
    LRUCache(int _capacity) 
    {
        size = 0;
        capacity = _capacity;
        head = new Node();
        tail = new Node();
        head->next = tail;
        tail->prev = head;
    }
    
    int get(int key) 
    {
        if(nodes.find(key) == nodes.end())
            return -1;
        
        auto node = nodes[key];
        moveToHead(node);

        return node->value;
    }
    
    void put(int key, int value) 
    {
        if(nodes.find(key) == nodes.end())
        {
            auto node = new Node(key, value);
            nodes[key] = node;
            addToHead(node);

            size++;
            if(size > capacity)
            {
                auto removedNode = removeLastNode();
                nodes.erase(nodes.find(removedNode->key));

                delete removedNode;
                size--;
            }
        }
        else
        {
            auto node = nodes[key];
            node->value = value;

            moveToHead(node);
        }
    }
};

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