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【leetcode刷题笔记】LRU Cache

2024-07-12 21:09:11   221人阅读

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：用双向链表实现一个LRU cache，这里自定义越靠近链表头的node，在越近的时间使用过。支持两个操作——get和set：

  1. get(key)：如果链表中有key对应的node，返回该node的值，并把node移到链表头部（最近使用过）；如果没有，返回-1；
  2. set(key,value)：如果链表中已经有key对应的node，修改对应node的值为value（但不需要把这个node放在头结点处）；如果链表中没有key对应的node，那么要新建node插入链表中，但此时要看链表是否还有空间。如果没有，就将尾部node（最少使用的node）删除，然后在头部插入新的node。

为了提高查找效率，这里使用一个hashMap存放<key,node>键值对，这样就可以在O(1)的时间判断cache中是否存在对应的key。

代码如下：

 1 public class LRUCache { 2     private class Node{ 3         int value; 4         int key; 5         Node before; 6         Node after; 7         public Node(int key,int value){ 8             this.value =http://www.mamicode.com/ value; 9             this.key = key;10             before = null;11             after = null;12         }13     }14     15     private int capacity;16     private HashMap<Integer, Node> map = new HashMap<Integer,Node>();17     private Node headNode = new Node(-1, -1);18     private Node tailNode = new Node(-1, -1);19     20     public LRUCache(int capacity) {21         this.capacity = capacity;22         headNode.after = tailNode;23         tailNode.before = headNode;24     }25     26     public void move_to_head(Node current){27         current.after = headNode.after;28         headNode.after = current;29         current.before = headNode;30         current.after.before = current;31     }32     public int get(int key) {33         //if we don‘t have this node34         if(!map.containsKey(key))35             return -1;36         37         //if we have this node, get it and move it to head38         Node current = map.get(key);39         current.before.after = current.after;40         current.after.before = current.before;41         move_to_head(current);42         43         return map.get(key).value;        44     }45     46     public void set(int key, int value) {47         //if we already have this node,just change its value48         if(get(key) != -1){49             map.get(key).value =http://www.mamicode.com/ value;50             return;51         }52         53         //if we indeed don‘t have this node, we first check capacity54         if(map.size() == capacity){55             map.remove(tailNode.before.key);56             tailNode.before.before.after = tailNode;57             tailNode.before = tailNode.before.before;58         }59         60         //now we are sure we have space for this new node,put it ahead of the list61         Node newNode = new Node(key, value);62         map.put(key, newNode);63         move_to_head(newNode);64     }65 }

在实现的时候，还设置了两个node：head和tail，真正的cache数据节点存放在二者之间。


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