试题
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
代码
最简单的就是get或put后到将值重新插入到hash里。
class LRUCache {
int capacity;
LinkedHashMap<Integer, Integer> hash;
public LRUCache(int capacity) {
this.capacity = capacity;
this.hash = new LinkedHashMap<>();
}
public int get(int key) {
Integer val = hash.get(key);
if(val == null){
return -1;
}
hash.remove(key);
hash.put(key, val);
return val;
}
public void put(int key, int value) {
hash.remove(key);
hash.put(key, value);
if(hash.size() > capacity){
hash.remove(hash.keySet().iterator().next());
}
}
}
/**
* Your LRUCache object will be instantiated and called as such:
* LRUCache obj = new LRUCache(capacity);
* int param_1 = obj.get(key);
* obj.put(key,value);
*/
public class LRUCache {
Map<Integer, Integer> map;
public LRUCache(int capacity) {
this.map = new LinkedHashMap<Integer, Integer>(128, 0.75f, true) {
protected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {
return size() > capacity;
}
};
}
public int get(int key) {
return map.getOrDefault(key, -1);
}
public void put(int key, int value) {
map.put(key, value);
}
}