LRU Cache

Question

• leetcode: LRU Cache | LeetCode OJ
• lintcode: (134) LRU Cache ### Problem Statement

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.

题解

Java

public class Solution {
private int capacity;
	private HashMap<Integer, Node> map = new HashMap<>();
	private Node head = new Node(-1, -1), tail = new Node(-1, -1);

	private class Node {
		Node prev, next;
		int val, key;

		public Node(int key, int val) {
			this.val = val;
			this.key = key;
			prev = null;
			next = null;
		}

// 		@Override
// 		public String toString() {
// 			return "(" + key + ", " + val + ") " + "last:"
// 					+ (prev == null ? "null" : "node");
// 		}
	}

	public Solution(int capacity) {
		this.capacity = capacity;
		tail.prev = head;
		head.next = tail;
	}

	public int get(int key) {
		if (!map.containsKey(key)) {
			return -1;
		}
		// remove current
		Node currentNode = map.get(key);
		currentNode.prev.next = currentNode.next;
		currentNode.next.prev = currentNode.prev;
		
		// move current to tail;
		moveToTail(currentNode);
		
		return map.get(key).val;
	}

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