LRU Cache

问题描述

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. 

 

解决思路

双向链表。

 

程序

public class LRUCache {
	class Node {
		int key;
		int value;
		Node prev;
		Node next;

		public Node(int key, int value) {
			this.key = key;
			this.value = value;
		}
	}

	int capacity;
	Node head;
	Node tail;
	HashMap<Integer, Node> map;

	public LRUCache(int capacity) {
		this.capacity = capacity;
		this.head = new Node(-1, -1);
		this.tail = new Node(-1, -1);
		head.next = tail;
		tail.prev = head;
		this.map = new HashMap<Integer, Node>();
	}

	public int get(int key) {
		if (map.containsKey(key)) {
			Node node = map.get(key);
			node.prev.next = node.next;
			node.next.prev = node.prev;
			moveToTail(node);
			return node.value;
		}
		return -1;
	}

	private void moveToTail(Node node) {
		tail.prev.next = node;
		node.prev = tail.prev;
		node.next = tail;
		tail.prev = node;
	}

	public void set(int key, int value) {
		if (get(key) != -1) {
			map.get(key).value = value;
			return;
		}
		if (map.size() == capacity) {
			Node first = head.next;
			map.remove(first.key);
			head.next = first.next;
			first.next.prev=head;
		}
		Node node = new Node(key, value);
		map.put(key, node);
		moveToTail(node);
	}
}

　　

转载于:https://www.cnblogs.com/harrygogo/p/4670439.html