leetcode 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

读题：模拟一个LRU（最近最少使用队列），实现它的put和get
分析：
1.无论是put还是get，只要是在队列中命中了此对象，就把它移到队列最前面，因为他是最近用到的。需要注意是如果对象在队尾，处理下边界情况
2.get没有命中，不需要做处理，put没有命中，需要新增此对象到队列中。
3.put方法需要注意队列空时初始化head，put时size+1，队列满时，删除队尾一个元素，对象移到对头的处理跟get一样

解题：
定义队列元素：

public class Node{
        Node pre;
        Node next;
        int key;
        int val;
    };

形成双向链表

ac代码：

class LRUCache {
    int capacity;
    int curCount = 0;
    Node head;
    Node tail;
    public LRUCache(int capacity) {
            this.capacity = capacity;
    }

    public int get(int key) {
        if(head == null){
            return -1;
        }
        else if(head.key == key){
            return head.val;
        }
        else{
            Node cur = head.next;
            while(cur != null){
                if(cur.key == key){//在队列中找到了key。下面把它变为head
                    if(tail == cur){//如果是尾部，先更换尾部
                        tail = cur.pre;
                    }
                    cur.pre.next = cur.next;
                    if(cur.next != null){
                         cur.next.pre = cur.pre ;
                    }
                    cur.next = head;
                    head.pre = cur;
                    head=cur;
                    return head.val;
                }
                else{
                    cur = cur.next;
                }
            }
            return -1;
        }
    }

    public void put(int key, int value) {
        if(head == null){
            head = new Node();
            head.key = key;
            head.val = value;
            curCount++;
            tail = head;
        }
        else if(head.key == key){
            head.val = value;
            return;
        }
        else{
            Node cur = head;
            while(cur != null){
                if(cur.key == key){//在队列中找到了key。下面把它变为head
                    if(tail == cur){//如果是尾部，先更换尾部
                        tail = cur.pre;
                    }
                    cur.pre.next = cur.next;
                    if(cur.next != null){
                         cur.next.pre = cur.pre ;
                    }
                    cur.next = head;
                    head.pre = cur;
                    head = cur;
                    head.val = value;
                    return;
                }
                else{
                    cur = cur.next;
                }
            }
            //队列不包含key，添加进去
            if(curCount == capacity){//队列满了,去掉最后一个
                if(capacity == 1){
                    head.key = key;
                    head.val = value;
                }
                else{
                    tail = tail.pre;
                    tail.next.pre = null;
                    tail.next = null;
                    Node temp = new Node();
                    temp.key = key;
                    temp.val = value;
                    temp.next = head;
                    head.pre = temp;
                    head = temp;
                }

            }
            else{//否则直接加到队头
                Node temp = new Node();
                temp.key = key;
                temp.val = value;
                temp.next = head;
                head.pre = temp;
                head = temp;
                curCount++;
            }

        }
    }
     class Node{
        Node pre;
        Node next;
         int key;
        int val;
    }
}

/**
 * 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);
 */

这是一个纯模拟的写法，没有做什么优化，也不是O(1)解法，待续。。。

今天想了一下，实现LRU应该没有其他取巧的办法，还是要通过双向链表来实现，于是想到了LinkedHashMap，查到LinkedHashMap是可以实现LRU模型的，那么就可以用LinkedHashMap来解这道题。

class LRUCache {
    class LRUMap<K,V> extends LinkedHashMap<K, V> implements Map<K, V>{
        int capacity;
        public LRUMap(int capacity) {
            super(16, (float) 0.75, true);
            this.capacity = capacity;
        }

        @Override
        protected boolean removeEldestEntry(Map.Entry<K,V> eldest){
            if(size() > capacity){
                return true;
            }
            return false;
        }
    }
    Map<Integer,Integer> lruMap ;
    public LRUCache(int capacity) {
            lruMap = new LRUMap<Integer,Integer>(capacity);

    }

    public int get(int key) {
        if(lruMap.containsKey(key)){
            return lruMap.get(key);
        }
        else {
            return -1;
        }
    }

    public void put(int key, int value) {
        lruMap.put(key, value);
    }
}

因为LinkedHashMap的get和put都是O(1)时间复杂度，因此这种解法是满足要求的。LinkedHashMap如何实现LRU见另一篇对LinkedHaspMap的源码阅读。