LRU Cache
Description
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. set(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.
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Hint
Train of Thought
LRU cache 机制: 调用get方法, 算used, 所以这个key会被放在最前 set一样的key, 会替换原来的值, 并被放在最前 超过capacity, 把最后的那个删了, 新的放在最前
The key to solve this problem is using a double linked list which enables us to quickly move nodes.
Code
public class LRUCache {
class Node{
Node next;
Node pre;
int key;
int value;
public Node(int key, int value){
pre = next = null;
this.key = key;
this.value = value;
}
}
private int size;
private HashMap<Integer, Node> map;
private Node head;
private Node tail;
public LRUCache(int capacity) {
head = new Node(0,0);
tail = new Node(0,0);
head.next = tail;
tail.pre = head;
map = new HashMap<Integer, Node>();
this.size = capacity;
}
public int get(int key) {
int res = -1;
if (map.containsKey(key)){
Node node = map.get(key);
res = node.value;
removeToTail(node);
}
return res;
}
public void set(int key, int value) {
if (map.containsKey(key)){
Node node = map.get(key);
node.value = value;
removeToTail(node);
} else {
if (map.size() == size){
map.remove(head.next.key);
deleteHead();
}
Node now = new Node(key,value);
now.pre = tail.pre;
tail.pre.next = now;
now.next = tail;
tail.pre = now;
map.put(key, now);
}
}
private void deleteHead(){
head.next.next.pre = head;
head.next = head.next.next;
}
private void removeToTail(Node node){
if (node == tail.pre){
return;
}
node.pre.next = node.next;
node.next.pre = node.pre;
node.pre = tail.pre;
tail.pre.next = node;
node.next = tail;
tail.pre = node;
}