Leetcode: 146. LRU Cache

URL：

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

描述：

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

解题思路

主要利用双向链表和hash map实现

代码

class LRUCache {        static class Node{        int key;        int val;        Node pre;        Node post;        Node(int key,int val){            this.key = key;            this.val = val;        }    }        Node head,tail;        int size;        int capacity;        Map<Integer,Node> map;        public LRUCache(int capacity) {            this.head = null;            this.tail = null;            this.size = 0;            this.capacity = capacity;            this.map = new HashMap<Integer, Node>();        }        public int get(int key) {            if(!map.containsKey(key)) {                return -1;            }            Node temp = map.get(key);            remove(temp);            add(temp);            return temp.val;        }        public void put(int key, int value) {            Node temp;            if(map.containsKey(key)){                temp = map.get(key);                remove(temp);                map.remove(key);                size--;            }            if(size == capacity){                Node h = head;                remove(h);                map.remove(h.key);                size--;            }            temp = new Node(key,value);            add(temp);            map.put(key,temp);            size++;        }        private void add(Node n){            if(head == null && tail == null) {                head = n;                tail = n;            }else {                tail.post = n;                n.pre = tail;                tail = n;            }        }        private void remove(Node n){            if(n==head && n==tail){                head = null;                tail = null;            }else if(n == head) {                head = head.post;                head.pre = null;            } else if(n == tail) {                tail = tail.pre;                tail.post = null;            }else{                n.pre.post = n.post;                n.post.pre = n.pre;            }            n.pre = null;            n.post = null;        }    }/** * 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); */