Day71 LRU 缓存机制

运用你所掌握的数据结构，设计和实现一个 LRU (最近最少使用) 缓存机制。实现 LRUCache 类

https://leetcode-cn.com/problems/lru-cache/

LRUCache(int capacity) 以正整数作为容量 capacity 初始化 LRU 缓存
int get(int key) 如果关键字 key 存在于缓存中，则返回关键字的值，否则返回 -1 。
void put(int key, int value) 如果关键字已经存在，则变更其数据值；如果关键字不存在，则插入该组「关键字-值」。
当缓存容量达到上限时，它应该在写入新数据之前删除最久未使用的数据值，从而为新的数据值留出空间。

进阶：你是否可以在 O(1) 时间复杂度内完成这两种操作？

示例1：

输入
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
输出
[null, null, null, 1, null, -1, null, -1, 3, 4]

解释
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // 缓存是 {1=1}
lRUCache.put(2, 2); // 缓存是 {1=1, 2=2}
lRUCache.get(1); // 返回 1
lRUCache.put(3, 3); // 该操作会使得关键字 2 作废，缓存是 {1=1, 3=3}
lRUCache.get(2); // 返回 -1 (未找到)
lRUCache.put(4, 4); // 该操作会使得关键字 1 作废，缓存是 {4=4, 3=3}
lRUCache.get(1); // 返回 -1 (未找到)
lRUCache.get(3); // 返回 3
lRUCache.get(4); // 返回 4

提示：

1 <= capacity <= 3000
0 <= key <= 3000
0 <= value <= 104
最多调用 3 * 10^4 次 get 和 put

Java解法

思路：

最简单的想法就是利用双端队列进行存储存入顺序，利用hashMap进行缓存存储
在存放时，key从栈顶入栈，如存在，更新位置；如溢出，利用队列移除队尾

在取出时：如存在，更新位置；调试修改之后是可用工作的，但效率不高超时了，用来复习下栈、队列操作

static class LRUCache {
    private final int size;
    HashMap mMap;
    Deque mDeque;

    public LRUCache(int capacity) {
        size = capacity;
        mMap = new HashMap<>();
        mDeque = new ArrayDeque<>();
    }

    public int get(int key) {
        if (mDeque.contains(key)) {
            refreshKeyOrder(key);
        }
        Integer integer = mMap.getOrDefault(key, -1);
        System.out.println("结果=="+integer);
        return integer;
    }

    public void put(int key, int value) {
        mMap.put(key, value);
        refreshKeyOrder(key);
        log();
    }


    private void refreshKeyOrder(int key) {
        if (mDeque.contains(key)) {
            //栈：头部进入，头部取出
            //更新使用栈的方法更新
            Stack stack = new Stack<>();
            while (!mDeque.isEmpty()) {
                Integer pop = mDeque.pop();
                if (pop != null) {
                    if (pop == key) {
                        while (!stack.isEmpty()) {
                            mDeque.push(stack.pop());
                        }
                        mDeque.push(pop);
                        break;
                    } else {
                        stack.push(pop);
                    }
                }
            }
        } else {
            if (mDeque.size() >= size) {
                //队列：队尾插入，队头取出
                //溢出使用队列的方法取出队尾
                Integer pop = mDeque.pollLast();//移除，也就是队尾
                mMap.remove(pop);
            }
            mDeque.push(key);
        }
    }

    public void log() {
        Deque integerDeque = new ArrayDeque<>();
        while (!mDeque.isEmpty()) {
            Integer poll = mDeque.poll();
            System.out.print(poll + "=" + mMap.getOrDefault(poll, -1)+";");
            integerDeque.offer(poll);
        }
        System.out.println();
        mDeque = integerDeque;
    }
}

优化计算，直接用链表存储顺序，并迁移

package sj.shimmer.algorithm.m4_2021;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;

/**
 * Created by SJ on 2021/4/8.
 */

class D71 {
    public static void main(String[] args) {
//        LRUCache lRUCache = new LRUCache(2);
//        lRUCache.put(1, 1); // 缓存是 {1=1}
//        lRUCache.log();
//        lRUCache.put(2, 2); // 缓存是 {1=1, 2=2}
//        lRUCache.log();
//        lRUCache.get(1);    // 返回 1
//        lRUCache.put(3, 3); // 该操作会使得关键字 2 作废，缓存是 {1=1, 3=3}
//        lRUCache.log();
//        lRUCache.get(2);    // 返回 -1 (未找到)
//        lRUCache.put(4, 4); // 该操作会使得关键字 1 作废，缓存是 {4=4, 3=3}
//        lRUCache.log();
//        lRUCache.get(1);    // 返回 -1 (未找到)
//        lRUCache.get(3);    // 返回 3
//        lRUCache.get(4);    // 返回 4


        String[] operation = { "LRUCache", "put", "put", "put", "put", "put", "get", "put", "get", "get", "put", "get", "put", "put", "put", "get", "put", "get", "get", "get", "get", "put", "put", "get", "get", "get", "put", "put", "get", "put", "get", "put", "get", "get", "get", "put", "put", "put", "get", "put", "get", "get", "put", "put", "get", "put", "put", "put", "put", "get", "put", "put", "get", "put", "put", "get", "put", "put", "put", "put", "put", "get", "put", "put", "get", "put", "get", "get", "get", "put", "get", "get", "put", "put", "put", "put", "get", "put", "put", "put", "put", "get", "get", "get", "put", "put", "put", "get", "put", "put", "put", "get", "put", "put", "put", "get", "get", "get", "put", "put", "put", "put", "get", "put", "put", "put", "put", "put", "put", "put" } ;
        int[][] num = new int[][]{{10}, {10, 13}, {3, 17}, {6, 11}, {10, 5}, {9, 10}, {13}, {2, 19}, {2}, {3}, {5, 25}, {8}, {9, 22}, {5, 5}, {1, 30}, {11}, {9, 12}, {7}, {5}, {8}, {9}, {4, 30}, {9, 3}, {9}, {10}, {10}, {6, 14}, {3, 1}, {3}, {10, 11}, {8}, {2, 14}, {1}, {5}, {4}, {11, 4}, {12, 24}, {5, 18}, {13}, {7, 23}, {8}, {12}, {3, 27}, {2, 12}, {5}, {2, 9}, {13, 4}, {8, 18}, {1, 7}, {6}, {9, 29}, {8, 21}, {5}, {6, 30}, {1, 12}, {10}, {4, 15}, {7, 22}, {11, 26}, {8, 17}, {9, 29}, {5}, {3, 4}, {11, 30}, {12}, {4, 29}, {3}, {9}, {6}, {3, 4}, {1}, {10}, {3, 29}, {10, 28}, {1, 20}, {11, 13}, {3}, {3, 12}, {3, 8}, {10, 9}, {3, 26}, {8}, {7}, {5}, {13, 17}, {2, 27}, {11, 15}, {12}, {9, 19}, {2, 15}, {3, 16}, {1}, {12, 17}, {9, 1}, {6, 19}, {4}, {5}, {5}, {8, 1}, {11, 7}, {5, 2}, {9, 28}, {1}, {2, 2}, {7, 4}, {4, 22}, {7, 24}, {9, 26}, {13, 28}, {11, 26}};
        createLru(operation, num);
    }

    private static void createLru(String[] operation, int[][] num) {
        LRUCache lRUCache = new LRUCache(1);
        for (int i = 0; i < operation.length; i++) {
            switch (operation[i]){
                case "LRUCache" :
                    lRUCache = new LRUCache(num[i][0]);
                    break;
                case "put":
                    lRUCache.put(num[i][0],num[i][1]);
                    break;
                case "get":
                    lRUCache.get(num[i][0]);
                    break;
                default:
                    break;
            }
        }
    }


    static class LRUCache {
        private final int size;
        HashMap mMap;
        List orderList;

        public LRUCache(int capacity) {
            size = capacity;
            mMap = new HashMap<>();
            orderList = new ArrayList<>();
        }

        public int get(int key) {
            if (orderList.indexOf(key)!=-1) {
                refreshKeyOrder(key);
            }
            return mMap.getOrDefault(key, -1);
        }
        public void put(int key, int value) {
            mMap.put(key, value);
            refreshKeyOrder(key);
        }
        private void refreshKeyOrder(int key) {
            int index = orderList.indexOf(key);
            if (index ==-1) {
                if (orderList.size()>=size) {
                    Integer remove = orderList.remove(size - 1);
                    mMap.remove(remove);
                }
            }else {
                orderList.remove(index);
            }
            orderList.add(0, key);
        }
    }

}

image

官方解

https://leetcode-cn.com/problems/lru-cache/solution/lruhuan-cun-ji-zhi-by-leetcode-solution/

哈希表 + 双向链表

提交解用了链表，没有在链表这块进行查找优化，直接用了系统API的indexof，实际上就是O(n)的遍历

官方解O(1)时间是通过哈希表来完成的，hash计算能快速定位位置，但不是绝对的O(1)

put和get的操作通过哈希表能快速定位，但双向链表不支持随机访问的，定位到实际上还是需要时间的

认识不够，只觉得这个解法好优秀、但不是很满足O(1)要求
- 时间复杂度：O(1)
- 空间复杂度：O(capacity)

Day71 LRU 缓存机制

运用你所掌握的数据结构，设计和实现一个 LRU (最近最少使用) 缓存机制 。实现 LRUCache 类

Java解法

官方解

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