leetcode 146. LRU Cache

文章目录

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.

The cache is initialized with a positive capacity.

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

  • 思路:用链表存内容,然后用散列表存映射关系

C++

class LRUCache {
public:
    int size;
    list<int> lru;
    unordered_map<int, list<int>::iterator> mp;
    unordered_map<int, int> kv;
    LRUCache(int capacity) {
        size=capacity;
    }
    void touch(int key){
        if(kv.count(key))
            lru.erase(mp[key]);
        lru.push_front(key);
        mp[key]=lru.begin();
    }
    int get(int key) {
        if(kv.count(key)==0)return -1;
        touch(key);
        return kv[key];
    }
    
    void put(int key, int value) {
        if(kv.size()==size&&kv.count(key)==0){// full
            mp.erase(lru.back());
            kv.erase(lru.back());
            lru.pop_back();
        }
        touch(key);
        kv[key]=value;
    }
};

/**
 * 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);
 */

Java

class LRUCache {
    private Map<Integer,ListNode> lru;
    ListNode head=null;
    ListNode tail=null;
    int c;
    public LRUCache(int _c) {
        this.c=_c;
        lru=new HashMap<Integer,ListNode>();
    }
    
    public int get(int key) {
        if(lru.containsKey(key)) {
            ListNode t=lru.get(key);
            t.update();
            return t.v;
        } else return -1;
    }
    
    public void put(int key, int value) {
        if(lru.containsKey(key)) {
            ListNode t=lru.get(key);
            t.v=value;
            t.update();
        } else {
            if(c==0)return;
            if(lru.size()==c){
                lru.remove(head.k);
                head.remove();
            }
            ListNode tmp=new ListNode(key,value);
            tmp.append();
            lru.put(key,tmp);
        }
    }
    class ListNode {
        int k,v;
        ListNode pre=null,nxt=null;
        public ListNode(int _k,int _v){
            this.k=_k;
            this.v=_v;
        }
        private void update() {
            if(tail==this)return;
            if(this!=head)
                this.pre.nxt=this.nxt;
            else 
                head=this.nxt;
            this.nxt.pre=this.pre;
            this.append();
        }
        private void append(){
            if(tail==null){
                head=this;
                tail=this;
            } else {
                this.nxt=null;
                this.pre=tail;
                tail.nxt=this;
                tail=this;
            }
        }
        private void remove(){
            if(tail==this) {
                head=null;
                tail=null;
            } else {
                head=this.nxt;
                head.pre=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);
 */
posted @ 2020-08-11 00:35  winechord  阅读(92)  评论(0编辑  收藏  举报