146. LRU Cache
2018-01-03 本文已影响0人
Jeanz
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
一刷
题解:
用double linkedlist, 最新加入/访问过的插到head, remove时从尾部
并且用map快速定位node
class LRUCache {
final Node head = new Node(0, 0);
final Node tail = new Node(0, 0);
final Map<Integer, Node> map;
final int capacity;
public LRUCache(int capacity) {
this.capacity = capacity;
map = new HashMap(capacity);
head.next = tail;
tail.prev = head;
}
public int get(int key) {
int res = -1;
if(map.containsKey(key)){
Node n = map.get(key);
remove(n);
insertToHead(n);
res = n.value;
}
return res;
}
public void put(int key, int value) {
if(map.containsKey(key)){
Node n = map.get(key);
remove(n);
n.value = value;
insertToHead(n);
} else {
if(map.size() == capacity){
map.remove(tail.prev.key);
remove(tail.prev);
}
Node n = new Node(key, value);
insertToHead(n);
map.put(key, n);
}
}
private void remove(Node n){
n.prev.next = n.next;
n.next.prev = n.prev;
}
private void insertToHead(Node n){
Node headNext = head.next;
head.next = n;
headNext.prev = n;
n.prev = head;
n.next = headNext;
}
class Node{
Node prev, next;
int key, value;
Node(int k, int v){
key = k;
value = v;
}
}
}
/**
* 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);
*/