Leetcode- LRUCache

来源：互联网发布：数据正规化编辑：程序博客网时间：2024/06/14 22:01

关键是搞懂题目（不知道LRUCache的只能自己google了）。

然后用双向链表来模拟cache被get和set。但是naive implementation会exceed time limit。所以最大的关键是用一个HashMap来记录key在链表中的位置，这样子每次查询是O(1)的时间，否则O(n)。

这个也是很经典的用Map来加速双向链表查询的思路（前提是key要唯一）。

import java.util.*;public class LRUCache {Map<Integer, DoubleLinkedListNode> hmap = new HashMap<Integer, DoubleLinkedListNode>();DoubleLinkedListNode head = null;DoubleLinkedListNode tail = null;int capa;public LRUCache(int capacity) {this.capa = capacity;}public int get(int key) {if (hmap.containsKey(key)) {DoubleLinkedListNode tnode = hmap.get(key);tnode = removeNode(tnode);pushFront(tnode);return tnode.value;}return -1;}public void set(int key, int value) {if (hmap.containsKey(key) == true) {DoubleLinkedListNode tnode = hmap.get(key);tnode.value = value;tnode = removeNode(tnode);pushFront(tnode);}else {DoubleLinkedListNode newNode = new DoubleLinkedListNode(key,value);if (hmap.size() == capa) {hmap.remove(tail.key);removeNode(tail);}pushFront(newNode);hmap.put(key, newNode);}}void pushFront(DoubleLinkedListNode node) {if (head == null) {head = tail = node;} else {node.next = head;head.prev = node;head = node;}return;}DoubleLinkedListNode removeNode(DoubleLinkedListNode node) {DoubleLinkedListNode prev = node.prev;DoubleLinkedListNode next = node.next;if (prev != null) {prev.next = next;} else {head = next;}if (next != null) {next.prev = prev;} else {tail = prev;}node.prev = null;node.next = null;return node;}class DoubleLinkedListNode {int key;int value;DoubleLinkedListNode prev;DoubleLinkedListNode next;public DoubleLinkedListNode(int key, int val) {this.key = key;this.value = val;prev = null;next = null;}public DoubleLinkedListNode() {value = 0;prev = null;next = null;}}}

0 0