【leetcode每日刷題】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.

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

使用鏈表 + 哈希表

使用鏈表模擬LRU的寄存器，每次從鏈表中取出某個元素之後，然後將該元素移動到鏈表的表頭，表頭代表最近訪問的元素。當put元素的時候，如果沒有達到容量，則直接put，如果達到容量，則去除鏈表的尾元素。

使用哈希表存儲key，存儲容量和裏面的key。

import java.util.HashMap;
import java.util.LinkedList;
class LRUCache {
    int capacity;
    LinkedList ageList;
    HashMap<Integer, Integer> cache;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        ageList = new LinkedList<Integer>();
        cache = new HashMap<>();
    }

    public int get(int key) {        
        Integer result = cache.get(key);
        if (result == null)
            return -1;
        ageList.remove(new Integer(key));
        ageList.addFirst(key);
        return result;       
    }

    public void put(int key, int value) {
        cache.put(key, value);
        if (cache.size() > capacity){            
            cache.remove(ageList.removeLast());            
        }
        ageList.remove(new Integer(key));
        ageList.addFirst(key);
    }
}

/**
 * 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);
 */