8.12 [LeetCode] 146 LRU Cache

Question

link

 

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

 

Stats

Adjusted Difficulty	4
Time to use	Very difficult

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

This is a difficult question, I can’t write the solution easily even after a month.

Solution

The solution is to use a Doubly-linked-list and a HashMap. Doing this allows O(1) search, remove and insert. A very nice and sophisticated data structure example, and very high frequency in interviews.

2 important things to note while coding:

1. We need 2 helper methods: removeNode() and setNodeAsHead().

   Because we reuse both methods for get() and set() methods.

2. Initialization of LRU

   We need 5 variables: capacity, current size(optional but good to have), hashmap, head, tail. Don't forget initialize tail.

3. Initialization of DoubleLinkedListNode (the key in List is the same as the key in map so we can reference from each other)

   This is easy, but do not forget about both key and value variable. We must use DoubleLinkedListNode.key when we want to delete tail.

Code

public class LRUCache {
    int size, capacity;
    HashMap<Integer, DoubleLink> map = new HashMap<Integer, DoubleLink>();
    DoubleLink head;
    DoubleLink tail;
    public LRUCache(int capacity) {
        this.capacity = capacity;
        this.size = 0;
        this.head = null;
        this.tail = null;
    }
    
    public void remove(DoubleLink node) {
        if(node == head && node == tail) {
            head = null;
            tail = null;
        }else if(node == tail) {
            tail.prev.next = null;
            tail = tail.prev;
        }else{
            node.prev.next = node.next;
            node.next.prev = node.prev;
        }
        node.prev = null;
        node.next = null;
    }
    
    public void setHead(DoubleLink node) {
        if(head != null) {
            head.prev = node;
            node.next = head;
            node.prev = null;
            head = node;
        }else { // first node inserted
            head = node;
            tail = node;
        }
    }
    
    public int get(int key) {
        if(! map.containsKey(key)) // key not found
            return -1;
        DoubleLink target = map.get(key);
        if(target != head) { // if it is already head we do not need to update it.
            remove(target);
            setHead(target);
        }
        return map.get(key).val;
    }
    
    public void set(int key, int value) {
        if(get(key) != -1) { // update old node
            DoubleLink node = map.get(key);
            node.val = value;
            if(node != head) {
                remove(node);
                setHead(node);
            }
        } else { // insert new node
            DoubleLink newHead = new DoubleLink(key, value);
            if(size == capacity) {
                map.remove(tail.key);
                remove(tail);
            }
            else
                size++;
            map.put(key, newHead);
            setHead(newHead);
        }
    }
    class DoubleLink { // DoubleLinkedList can delete a node moer easily cause you can use node.prev to connect node.next.
        int val;
        int key;
        DoubleLink next;
        DoubleLink prev;
        public DoubleLink(int k, int v) {
            this.key = k;
            this.val = v;
        }
    }
}