LeetCode | LRU Cache

题目

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

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

投机取巧的方法就是直接利用Java类库中提供的LinkedHashMap类，该类的作用之一就是辅助实现LRU Cache。

如果自己实现的话，可以参照LinkedHashMap的实现原理，用传统的HashMap实现O(1)时间复杂度的存取，再用一个双向链表纪录访问的先后顺序，最后访问的放在表尾，最久没被访问的放在表首，如果达到容量上限需要删除Entry时，根据表首元素，分别在HashMap和双向链表中进行删除。

代码

import java.util.LinkedHashMap;
import java.util.Map;

public class LRUCache {
	private int capacity;
	private Map<Integer, Integer> map;

	public LRUCache(int c) {
		this.capacity = c;
		this.map = new LinkedHashMap<Integer, Integer>(capacity, 0.75f, true) {
			private static final long serialVersionUID = 7499786237341382662L;

			protected boolean removeEldestEntry(
					Map.Entry<Integer, Integer> eldest) {
				return size() > capacity;
			}
		};
	}

	public int get(int key) {
		if (!map.containsKey(key)) {
			return -1;
		}
		return map.get(key);
	}

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

LeetCode | LRU Cache

原文：http://blog.csdn.net/perfect8886/article/details/18967197