146. LRU Cache

题目

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.

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

思路

设计一个LRU算法。

操作系统中进行内存管理中时采用一些页面置换算法,如LRU、LFU和FIFO等。其中LRU应用较为广泛。LRU的全称是Least Recently Used,即最近最少使用算法。当访问数据时,如缓存中有数据,则将该数据移动至链表的顶端;没有该数据则在顶端加入该数据,并移除链表中的最不常使用的数据。

这题应该用双链表来实现,使用Queue的话会超时因为Queue的查找使用了O(n),双链表可以在O(1)完成get(key)`put(key)`操作。

  • Cache 使用一个HashMap来达到O(1)的查找效果。
  • Get 用一个双链表,自定义一个Node,里面有key,value,preNode,nextNode。
    • 假如miss,返回-1(rise exception?)
    • 假如 hit:
      • 记录当前节点的value
      • 在双链表中删除这个Node
      • 链表头部加入这个Node
  • Set
    • 假如miss:
      • 假如List.Size()<Capacity:
        • 添加新节点到队头
        • 增加新节点到map
      • 假如 List.Size()>Capacity:
        • 删掉队尾节点
        • 增加新节点到map
        • 添加新节点到队头
    • 假如 hit:
      • 更新vaule值
      • 节点移到队头

代码

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class LRUCache {
class Node {
int key;
int value;
Node pre;
Node next;

public Node(int key, int value) {
this.key = key;
this.value = value;
}
}
HashMap<Integer, Node> map;
int capacity, count;
Node head, tail;

public LRUCache(int capacity) {
this.capacity = capacity;
map = new HashMap<Integer, Node>();
head = new Node(0,0);
tail = new Node(0,0);
head.next = tail;
head.pre = null;
tail.pre = head;
tail.next = null;
count = 0;
}

private void removeNode(Node node) {
node.pre.next = node.next;
node.next.pre = node.pre;
}

private void addNodeToHead(Node node) {
node.next = head.next;
node.next.pre = node;

node.pre = head;
head.next = node;

}

public int get(int key) {
Node node = map.get(key);
if(node !=null) {
int result = node.value;
removeNode(node);
addNodeToHead(node);
return result;
}
else {
return -1;
}
}

public void put(int key, int value) {
if(map.get(key)!=null) {
Node node = map.get(key);
node.value = value;
removeNode(node);
addNodeToHead(node);
} else {
Node node = new Node(key,value);
map.put(key,node);
if(count<capacity) {
count++;
} else {
map.remove(tail.pre.key);
removeNode(tail.pre);
}
addNodeToHead(node);
}
}
}
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```


Java里面有个数据结构叫`LinkedHashMap`,刚好可以满足LRU算法,不过面试的时候应该不会让你用吧,不然对我们这种C++选手太不公平了!

```java
//Java Solution
public class LRUCache {
private LinkedHashMap<Integer, Integer> map;
private final int CAPACITY;
public LRUCache(int capacity) {
CAPACITY = capacity;
map = new LinkedHashMap<Integer, Integer>(capacity, 0.75f, true){
protected boolean removeEldestEntry(Map.Entry eldest) {
return size() > CAPACITY;
}
};
}
public int get(int key) {
return map.getOrDefault(key, -1);
}
public void put(int key, int value) {
map.put(key, value);
}
}
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