
题目描述
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
解决思路
使用双向链表+哈希表
源码
import java.util.Hashtable;
class LRUCache {
class Node{
int key;
int value;
Node next=null;
Node pre=null;
Node(int key,int value){
this.key=key;
this.value=value;
}
}
private void removeNode(Node node){
node.next.pre=node.pre;
node.pre.next=node.next;
}
private void addNode(Node node){
head.next.pre=node;
node.next=head.next;
head.next=node;
node.pre=head;
}
private void moveToHead(Node node){
removeNode(node);
addNode(node);
}
private Node head,tail;
private Hashtable<Integer,Node>cache;
private int count;
private int capacity;
public LRUCache(int capacity) {
cache=new Hashtable<>(capacity);
head=new Node(-1,-1);
tail=new Node(-1,-1);
head.next=tail;
tail.pre=head;
this.capacity=capacity;
count=0;
}
public int get(int key) {
Node node=cache.get(key);
if(node==null){
return -1;
}else{
moveToHead(node);
return node.value;
}
}
public void put(int key, int value) {
Node node=cache.get(key);
if(node!=null){
node.value=value;
moveToHead(node);
return;
}else{
node=new Node(key,value);
addNode(node);
if(count==capacity){
cache.remove(tail.pre.key);
removeNode(tail.pre);
}else{
count++;
}
cache.put(key,node);
}
}
}




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