146.LRU-Cache

146. LRU Cache

题目地址

https://www.lintcode.com/problem/lru-cache

https://leetcode.com/problems/lru-cache/

题目描述

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.
Finally, you need to return the data from each get.

代码

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

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

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

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

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

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