zoukankan      html  css  js  c++  java
  • 【LeetCode】LRU Cache

    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.


    import java.util.HashMap;
    import java.util.LinkedList;
    
    
    
    class ListNode{
    	int key, value;
    	ListNode next, prev;
    	public ListNode()
    	{
    		this.key = this.value = 0;
    		next = prev = null;
    	}
    	public ListNode(int key, int val)
    	{
    		this.key = key;
    		this.value = val;
    		next = prev = null;
    	}
    }
    
    public class LRUCache {
    	private ListNode head,tail;
    	private HashMap<Integer, ListNode> hashmap;
    	private int size, capacity;
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    
    	}
    	
    	public LRUCache(int capacity) {
            this.size = 0;
            this.capacity = capacity;
            head = new ListNode();
            tail = new ListNode();
            head.next = tail;
            tail.prev = head;
            hashmap = new HashMap<Integer, ListNode>();
        }
        
        public int get(int key) {
            if (hashmap.containsKey(key)) {
    			ListNode cur = hashmap.get(key);
    			//detach the node in the list
    			cur.prev.next = cur.next;
    			cur.next.prev = cur.prev;
    			//set the visting node be the firstNode.
    			ListNode tmp = head.next;
    			head.next = cur;
    			cur.prev = head;
    			cur.next = tmp;
    			tmp.prev = cur;
    			return cur.value;
    		}
            else return -1;
        }
        
        public void set(int key, int value) {
            if (hashmap.containsKey(key)) {
    			ListNode cur = hashmap.get(key);
    			cur.value = value;
    			//detach
    			cur.prev.next = cur.next;
    			cur.next.prev = cur.prev;
    			//set the visting node be the firstNode.
    			ListNode tmp = head.next;
    			head.next = cur;
    			cur.prev = head;
    			cur.next = tmp;
    			tmp.prev = cur;
    		}
            else {
    			if (size == capacity) {
    				ListNode last = tail.prev;
    				hashmap.remove(last.key);
    				tail = last;
    				last = null;
    				size--;
    			}
    			ListNode cur = new ListNode(key, value);
    			hashmap.put(key, cur);
    			//set the new node be the firstNode.
    			ListNode tmp = head.next;
    			head.next = cur;
    			cur.prev = head;
    			cur.next = tmp;
    			tmp.prev = cur;
    			size++;
    		}
        }
    
    }
    


    版权声明:本文博主原创文章。博客,未经同意不得转载。

  • 相关阅读:
    愤怒
    Eclipse的调试功能的10个小窍门
    PL/SQL之基础篇
    PL/SQL之高级篇
    luogu P1015 回文数
    Noip2011 提高组 Day1 T3 Mayan游戏
    各种各样的——玄学卡常技巧
    北京清北 综合强化班 Day5
    [UVA12003] Array Transformer(分块,二分,暴力)
    [POJ3468] A Simple Problem with Integers(分块)
  • 原文地址:https://www.cnblogs.com/lcchuguo/p/4777627.html
Copyright © 2011-2022 走看看