• 146. LRU Cache


    Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

    get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
    put(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.

    Follow up:
    Could you do both operations in O(1) time complexity?

    Example:

    LRUCache cache = new LRUCache( 2 /* capacity */ );
    
    cache.put(1, 1);
    cache.put(2, 2);
    cache.get(1);       // returns 1
    cache.put(3, 3);    // evicts key 2
    cache.get(2);       // returns -1 (not found)
    cache.put(4, 4);    // evicts key 1
    cache.get(1);       // returns -1 (not found)
    cache.get(3);       // returns 3
    cache.get(4);       // returns 4

    HashMap + Double LinkedList

    首先定义一个Node类,用来构建double linked list,包括key, val, next, prev。两个参数count, capacity,count用来给cache里的元素计数。LRUcache初始化的时候,还要定义两个dummy Node,head和tail,方便取和放链表首(head.next)、尾(tail.prev)的元素。

    get:如果map中存在该key,从map中取出node,保存这个node对应的value,在链表中删掉这个节点,再把这个节点插入到链表头部。如果map中不存在该key,返回-1。

    put:根据map中有无该key,分两种情况

    1. map中无key。根据新的(key, value)新建一个node,把node放进map里。如果没有超过容量,把当前node放到double linked list的头部,并且count增加1;如果超过容量,删除链表尾部的节点(注意除了链表里,map里也要删除!),并把当前node插入到头部。

    2. map中有key。从map中取出node,并赋值为当前value,在链表中删掉这个节点,再把这个节点插入到链表头部。

    删除节点 和 把节点插入到链表头部 可以用两个辅助函数来表示。

    时间:O(1),空间:O(N)

    class LRUCache {
        class Node {
            int key, val;
            Node next, prev;
            public Node(int key, int val) {
                this.key = key;
                this.val = val;
            }
        }
        HashMap<Integer, Node> map;
        Node head, tail;
        int count, capacity;
    
        public LRUCache(int capacity) {
            this.capacity = capacity;
            count = 0;
            map = new HashMap<>();
            head = new Node(0, 0);
            tail = new Node(0, 0);
            head.next = tail;
            head.prev = null;
            tail.next = null;
            tail.prev = head;
        }
        
        private void deleteNode(Node node) {
            node.prev.next = node.next;
            node.next.prev = node.prev;
        }
        
        private void addToHead(Node node) {
            head.next.prev = node;
            node.next = head.next;
            head.next = node;
            node.prev = head;
        }
        
        public int get(int key) {
            if(map.containsKey(key)) {
                Node node = map.get(key);
                int val = node.val;
                deleteNode(node);
                addToHead(node);
                return val;
            }
            return -1;
        }
        
        public void put(int key, int value) {
            if(map.containsKey(key)) {
                Node node = map.get(key);
                node.val = value;
                deleteNode(node);
                addToHead(node);
            } else {
                Node node = new Node(key, value);
                map.put(key, node);
                if(count < capacity) {
                    addToHead(node);
                    count++;
                } else {
                    map.remove(tail.prev.key);
                    deleteNode(tail.prev);
                    addToHead(node);
                }
            }
        }
    }
    
    /**
     * Your LRUCache object will be instantiated and called as such:
     * LRUCache obj = new LRUCache(capacity);
     * int param_1 = obj.get(key);
     * obj.put(key,value);
     */
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10048111.html
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