146. LRU Cache (Hard)
https://leetcode.com/problems/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.
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
Solutions
1.
class LRUCache {
int capacity = 0;
int version = 0;
Map<Integer, Integer> versionMap = new HashMap<>();
Map<Integer, Integer> lruCache = new HashMap<>();
Queue<Integer> versionQueue = new PriorityQueue<>(Comparator.comparingInt(o -> versionMap.get(o.intValue())));
public LRUCache(int capacity) {
this.capacity = capacity;
}
public int get(int key) {
if (!lruCache.containsKey(key)) {
return -1;
}
updateVersion(key);
return lruCache.get(key);
}
public void put(int key, int value) {
if (lruCache.containsKey(key)) {
lruCache.put(key, value);
updateVersion(key);
return;
}
// if exceeds the capacity of queue, remove least recent used element at first
if (versionMap.size() >= capacity) {
removeLRU();
}
version++;
lruCache.put(key, value);
versionMap.put(key, version);
versionQueue.add(key);
}
private void updateVersion(Integer key) {
version++;
// update version
versionMap.put(key, version);
// resort version queue
versionQueue.remove(key);
versionQueue.add(key);
}
private void removeLRU() {
int k = versionQueue.poll();
lruCache.remove(k);
versionMap.remove(k);
versionQueue.remove(k);
}
}