M
tags: Design, Linked List, Hash Table, Doubly Linked List
time: O(1)
space: O(1)
#### Double Linked List
- 用了一个特别的双向的ListNode，有了head和tail，这样就大大加快了速度
- 主要加快的就是那个‘更新排位’的过程，找到item hashmap O(1), 做减法换位也都是O(1)
- Overall O(1)
- 巧妙点
- 1. head和tail特别巧妙：除掉头和尾，和加上头和尾，都O(1)
- 2. remove node: 把node.pre和node.next 连起来, node就自然而然的断开不要了
- 一旦知道怎么解决了，就不是很特别，并不是难写的算法
- moveToHead()
- insertHead()
- remove()
#### Deque, less efficient
- Instead of building `Double Linked List`, utilize Java `Deque queue = new LinkedList<>()`
- works but problem: `queue.remove(E)` is O(n)
- time: O(1) on average but much slower
```
/*
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
*/
/*
It looks like we need to do some design, according to (http://www.cnblogs.com/yuzhangcmu/p/4113462.html). Though, the solution's concept is quite similar as attempt1.
1. The design uses HashMap, and 2-way LinkedList. Map
2. Use two dummy node: head and tail. When add/remove nodes, we are add/remove nodes in between head and tail.
So. head.next is our real 1st element
andd tail.pre is our real last element
Note:
Be careful: when removing a node due to capacity issue, remember to remove both 1st node(head.next) and the corresponding entry in the map: map.remove(head.next.key)
*/
//Use double linked list to store value.
//Store key in hashmap to find node easily
//Functions: insert in front, remove node,
public class LRUCache {
class DoubleLinkedListNode {
int key, val;
DoubleLinkedListNode next,prev;
public DoubleLinkedListNode(int key, int val){
this.key = key;
this.val = val;
}
}
public int capacity;
public HashMap map;
public DoubleLinkedListNode head, tail;
public LRUCache(int capacity) {
this.capacity = capacity;
this.map = new HashMap<>();
this.head = new DoubleLinkedListNode(-1, -1);
this.tail = new DoubleLinkedListNode(-1, -1);
head.next = tail;
head.prev = tail;
tail.next = head;
tail.prev = head;
}
public int get(int key) {
if(!map.containsKey(key)) return -1;
DoubleLinkedListNode node = map.get(key);
moveToHead(node);
return node.val;
}
public void put(int key, int value) {
if (map.containsKey(key)) {
map.get(key).val = value;
moveToHead(map.get(key));
} else {
DoubleLinkedListNode node = new DoubleLinkedListNode(key, value);
if (map.size() >= this.capacity) {
DoubleLinkedListNode rm = tail.prev;
remove(rm);
map.remove(rm.key);
}
insertHead(node);
map.put(key, node);
}
}
public void moveToHead(DoubleLinkedListNode node) {
remove(node);
insertHead(node);
}
//Helper functions
//Put node to front, where the latest item is at.
public void insertHead(DoubleLinkedListNode node) {
DoubleLinkedListNode next = head.next;
head.next = node;
node.prev = head;
node.next = next;
next.prev = node;
}
// Find front and end, link them.
public void remove(DoubleLinkedListNode node) {
DoubleLinkedListNode front = node.prev;
DoubleLinkedListNode end = node.next;
front.next = end;
end.prev = front;
}
}
// Deque Less Efficient:
// Deque remove() loops over the list to remove, so it's inefficient
public class LRUCache {
class Node {
int key, val;
public Node(int key, int val){
this.key = key;
this.val = val;
}
}
public int capacity;
public HashMap map;
public Deque queue;
public LRUCache(int capacity) {
this.capacity = capacity;
this.map = new HashMap<>();
this.queue = new LinkedList<>();
}
public int get(int key) {
if(!map.containsKey(key)) return -1;
Node node = map.get(key);
moveToHead(node);
return node.val;
}
public void put(int key, int value) {
if (map.containsKey(key)) {
map.get(key).val = value;
moveToHead(map.get(key));
} else {
Node node = new Node(key, value);
if (map.size() >= this.capacity) {
Node rm = queue.pollLast();
map.remove(rm.key);
}
queue.offerFirst(node);
map.put(key, node);
}
}
public void moveToHead(Node node) {
queue.remove(node);
queue.offerFirst(node);
}
}
```