LeetCode解题报告:LRU Cache
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.
思路:Java的LinkedHashMap可以实现最近最少使用(LRU)的次序。类似于HashMap。详见《JAVA编程思想(第4版)》P487.
题目要求是一个固定大小的cache,因此需要一个变量maxCapacity来记录容量大小,用LinkedHashMap存储数据。在添加数据set()方法时,判断一下是否达到maxCapacity,如果cache已经满了,remove掉最长时间不使用的数据,然后put进新的数据。
注意:HashMap,LinkedHashMap,TreeMap的区别,详细看看StackOverFlow
LinkedHashMap最常用的是LRU cache的实现。
如果用C++实现的话, hashmap + 双向链表:用 双向链表记录value 用hashmap记录 key值在链表中的位置(指针)。
unordered_map<int, list<CacheNode>:: iterator> cacheMap;
题解:
import java.util.LinkedHashMap; public class LRUCache { LinkedHashMap<Integer, Integer> linkedmap; int maxCapacity; public LRUCache(int capacity) { this.maxCapacity = capacity; this.linkedmap = new LinkedHashMap<Integer, Integer>(capacity, 1f, true); } public int get(int key) { if (linkedmap.containsKey(key)) return linkedmap.get(key); else return -1; } public void set(int key, int value) { int size = linkedmap.size(); if ((size < maxCapacity) || (linkedmap.containsKey(key))) { linkedmap.put(key, value); } else if (size >= maxCapacity) { Iterator<Integer> it = linkedmap.keySet().iterator();//iterator method is superior the toArray(T[] a) method. linkedmap.remove(it.next()); linkedmap.put(key, value); } } }
结题遇到的问题:
1.下面这段代码提交的时候超时了。
import java.util.LinkedHashMap; public class LRUCache { LinkedHashMap<Integer, Integer> linkedmap; int maxCapacity; public LRUCache(int capacity) { this.maxCapacity = capacity; this.linkedmap = new LinkedHashMap<Integer, Integer>(capacity, 1f, true); } public int get(int key) { if (linkedmap.containsKey(key)) return linkedmap.get(key); else return -1; } public void set(int key, int value) { int size = linkedmap.size(); if ((size < maxCapacity) || (linkedmap.containsKey(key))) { linkedmap.put(key, value); } else if (size >= maxCapacity) { Integer[] keyArray = linkedmap.keySet().toArray(new Integer[0]);//这是超时的代码,采用Iterator不会超时。 linkedmap.remove(keyArray[0]); linkedmap.put(key, value); } } }
2.Roger自己实现LinkedHashMap的功能,采用双向链表和哈希表。效率略低于LinkedHashMap.(644ms>548ms)
import java.util.HashMap; public class LRUCache { private HashMap<Integer, Entry<Integer>> index; private UDFList<Integer> data; public LRUCache(int capacity) { index = new HashMap<Integer, Entry<Integer>>(capacity); data = new UDFList<Integer>(capacity); } public int get(int key) { if (!isExist(key)) { index.remove(key); return -1; } if (!index.get(key).equals(data.head)) { Entry<Integer> nodePtr = data.adjust(index.get(key)); index.put(key, nodePtr); } return index.get(key).element; } public void set(int key, int value) { if (isExist(key)) { data.remove(index.get(key)); } index.put(key, data.push(value)); } private boolean isExist(int key) { if (index.get(key) == null) { return false; } if (index.get(key).element == null) { return false; } return true; } public class UDFList<E> { public Entry<E> head; public Entry<E> tail; public final int size; public int length = 0; public UDFList(int size) { head = new Entry<E>(null, null, null); tail = head; this.size = size; } public Entry<E> adjust(Entry<E> node) { if (node.equals(tail)) { tail = tail.previous; tail.next = null; node.previous = null; } else if (node.equals(head)) { node = null; return head; } else { node.previous.next = node.next; node.next.previous = node.previous; } head.previous = node; node.next = head; head = node; node = null; return head; } public Entry<E> push(E e) { Entry<E> newNode = new Entry<E>(e, null, null); if (length == 0) { head = newNode; tail = head; } else { head.previous = newNode; newNode.next = head; head = newNode; } if (length == size) { remove(tail); } length++; return head; } public void remove(Entry<E> node) { if (node == null) return; node.element = null; if (node.equals(head)) { head = head.next; } else if (node.equals(tail)) { tail = tail.previous; tail.next = null; } else { node.previous.next = node.next; node.next.previous = node.previous; } node = null; length--; } } public class Entry<E> { public E element; public Entry<E> previous; public Entry<E> next; public Entry(E element, Entry<E> next, Entry<E> previous) { this.element = element; this.next = next; this.previous = previous; } } }
posted on 2014-06-13 16:27 BYRHuangQiang 阅读(534) 评论(0) 编辑 收藏 举报