LRU Cache
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(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.
The cache is initialized with a positive capacity.
Follow up:
Could you do both operations in O(1) time complexity?
Example:
LRUCache cache = new LRUCache( 2 );
cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // returns 1
cache.put(3, 3); // evicts key 2
cache.get(2); // returns -1 (not found)
cache.put(4, 4); // evicts key 1
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4
Solution
The shortcut is to use LinkedHashMap with capacity 2. We also need to override
removeEldestEntry
@Override
protected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {
return size() > capacity;
}
The LinkedHashMap implements the eviction with a double linked list. The hashmap maps the key to a double linked list node. Whenever a node is added, it is added after the head node. If the size is bigger
than capacity, a node is removed from tail. A get operation remove a linked list node, put it at the head.
import java.util.*;
class Solution {
public static void main(String[] args) {
LRUCache cache = new LRUCache(2);
cache.put(1, 1);
cache.put(2, 2);
System.out.println(cache.get(1)); // returns 1
cache.put(3, 3); // evicts key 2
System.out.println(cache.get(2)); // returns -1 (not found)
cache.put(4, 4); // evicts key 1
System.out.println(cache.get(1)); // returns -1 (not found)
System.out.println(cache.get(3)); // returns 3
System.out.println(cache.get(4)); // returns 4
}
static class LRUCache {
private int capacity = 0;
private Node head;
private Node tail;
private int size;
private Map<Integer, Node> map = new HashMap<>();
public LRUCache(int capacity) {
this.capacity = capacity;
head = new Node();
tail = new Node();
head.next = tail;
tail.next = head;
}
public int get(Integer key) {
Node node = map.get(key);
if(node != null) {
removeNode(node);
addNode(node);
return node.val;
} else {
return -1;
}
}
public void put(Integer key, Integer val) {
Node node = new Node();
node.key = key;
node.val = val;
addNode(node);
size++;
map.put(key, node);
if(size > capacity) {
Node tail = removeTail();
size--;
map.remove(tail.key);
}
}
private void removeNode(Node node) {
Node pre = node.pre;
Node next = node.next;
pre.next = next;
next.pre = pre;
}
// head -> node
// head -> newNode -> node
private void addNode(Node node) {
Node next = head.next;
node.next = next;
node.pre = head;
head.next = node;
next.pre = node;
}
// noden -> node -> tail
private Node removeTail() {
Node node = tail.pre;
Node pre = node.pre;
pre.next = tail;
tail.pre = pre;
return node;
}
}
static class Node {
int key;
int val;
Node pre;
Node next;
}
}
All the operations has cost O(1), the space is O(capacity), the extra space came from the double linked list with at most size equal the capacity.
