146. LRU 缓存机制
Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.
Implement the LRUCache
class:
LRUCache(int capacity)
Initialize the LRU cache with positive sizecapacity
.int get(int key)
Return the value of thekey
if the key exists, otherwise return-1
.void put(int key, int value)
Update the value of thekey
if thekey
exists. Otherwise, add thekey-value
pair to the cache. If the number of keys exceeds thecapacity
from this operation, evict the least recently used key.
Follow up:
Could you do get
and put
in O(1)
time complexity?
Example 1:
Input ["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"] [[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]] Output [null, null, null, 1, null, -1, null, -1, 3, 4] Explanation LRUCache lRUCache = new LRUCache(2); lRUCache.put(1, 1); // cache is {1=1} lRUCache.put(2, 2); // cache is {1=1, 2=2} lRUCache.get(1); // return 1 lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3} lRUCache.get(2); // returns -1 (not found) lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3} lRUCache.get(1); // return -1 (not found) lRUCache.get(3); // return 3 lRUCache.get(4); // return 4
Constraints:
1 <= capacity <= 3000
0 <= key <= 3000
0 <= value <= 104
- At most
3 * 104
calls will be made toget
andput
.
LRU缓存器。
实现这个缓存器,时间复杂度要求是O(1)。
LRU缓存器是在一定的容量范围内存了一些node,按照上一次被访问的时间由新到旧排列,越近被访问的node应该排得越靠前。有几个函数需要实现。
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.
时间O(1) - required
空间O(n) - hashmap + DLL
思路:
双向链表(DLL) + hashmap。
因为题意要求了时间复杂度必须是O(1)所以只有hashmap才能满足这个时间复杂度。
至于为什么是DLL而不是单链表,则是为了node之间的移动方便,也是为了添加删除节点的时候能更加高效。
class LRUCache { /** * map<K,Node> */ private HashMap<Integer,Node> keyNodeMap; /** * 自己组织的双链表 */ private NodeDoubleLinkedList nodeList; /** * 缓存容量 */ private Integer capacity; /*** * 构造函数初始化结构 */ public LRUCache(Integer capacity) { if (capacity < 1) { throw new RuntimeException("capacity should be more than 0"); } keyNodeMap = new HashMap<>(); nodeList = new NodeDoubleLinkedList(); this.capacity = capacity; } /** * get(k) */ public int get(int k) { //是否在Map中 if (keyNodeMap.containsKey(k)) { //通过K 定位到Node Node res = keyNodeMap.get(k); //刷新nodeList this.nodeList.moveNodeToTail(res); return res.v; } return -1; } /** * set(k,v) */ public void put(int k, int v) { //如已存在于表中,则刷新节点,并更新node.v if (keyNodeMap.containsKey(k)) { Node res = keyNodeMap.get(k); res.v = v; this.nodeList.moveNodeToTail(res); } else { //不存在则加入 Node node = new Node(k, v); //放入表中 keyNodeMap.put(k, node); //加入到链表上 this.nodeList.addNode(node); //如果容量超限,则移除头节点,交从表中删除 if (keyNodeMap.size() == capacity + 1) { Node removeNode = this.nodeList.removeHead(); keyNodeMap.remove(removeNode.k); } } } public class Node { public int k; public int v; private Node pre; private Node next; public Node(int k, int v) { this.k = k; this.v = v; } } /** * 先准备双链表的api */ public class NodeDoubleLinkedList { private Node head; private Node tail; public NodeDoubleLinkedList() { this.head = null; this.tail = null; } /** * 添加节点的操作,添加到结尾 */ public void addNode(Node node) { //如等待加的节点为空,直接返回 if (node == null) { return; } //如果head为null,则挂在头节点 if (head == null) { head = node; tail = node; } //否则挂在结尾的节点 tail.next=node; node.pre=tail; node.next=null; tail=node; } /** * 先保证这个节点存在的情况下考虑此问题 * 如果移动节点到尾部:先断联,再加到结尾节点 */ public void moveNodeToTail(Node node) { //如果要移动的是尾节点,则不用操作 if (this.tail == node) { return; } //先断联 //如果要移动的是头节点,更改头节点为老头的下一个节点 if(head==node){ head=head.next; head.pre=null; }else{//如果不是头尾节点,则是中间节点,先断联 node.pre.next=node.next; node.next.pre=node.pre; } //再移到尾节点 tail.next=node; node.pre=tail; node.next=null; tail=node; } /** * 移除节点 */ public Node removeHead() { //你要操作头节点,则必须先检查head的状态 //如果head null if (head == null) { return null; } //记录一下老头 Node res = head; //如果头节点是head==tail if (head == tail) { head = null; tail = null; } else {//链表节点个数>1 head = res.next; head.pre = null; res.next = null;//help gc } return res; } } }