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 /* capacity */ );
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
class LRUCache {
Node head = new Node(0, 0), tail = new Node(0, 0);
HashMap<Integer, Node> map = new HashMap();
int capacity;
public LRUCache(int capacity) {
this.capacity = capacity;
head.next = tail;
tail.prev = head;
}
public int get(int key) {
if (map.containsKey(key)) {
Node node = map.get(key);
remove(node);
insert(node);
return node.value;
} else {
return -1;
}
}
public void put(int key, int value) {
if (map.containsKey(key)) {
remove(map.get(key));
}
if (map.size() == capacity) {
remove(tail.prev);
}
insert(new Node(key, value));
}
private void remove(Node node) {
map.remove(node.key);
node.prev.next = node.next;
node.next.prev = node.prev;
}
private void insert(Node node) {
map.put(node.key, node);
Node headNext = head.next;
head.next = node;
node.prev = head;
headNext.prev = node;
node.next = headNext;
}
class Node {
Node prev, next;
int key, value;
Node(int key, int value) {
this.key = key;
this.value = value;
}
}
}
Last updated