LRU 缓存

Problem

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

• LRUCache(int capacity) Initialize the LRU cache with positive size capacity.
• int get(int key) Return the value of the key if the key exists, otherwise return -1.
• void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

The functions get and put must each run in O(1) average time complexity.

Example 1:

Input: ["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"] [[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]] Output: [null, null, null, 1, null, -1, null, -1, 3, 4]

Explanation:

• LRUCache lRUCache = new LRUCache(2);
• lRUCache.put(1, 1); // cache is {1=1}
• lRUCache.put(2, 2); // cache is {1=1, 2=2}
• lRUCache.get(1); // return 1
• lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
• lRUCache.get(2); // returns -1 (not found)
• lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
• lRUCache.get(1); // return -1 (not found)
• lRUCache.get(3); // return 3
• lRUCache.get(4); // return 4

Constraints:

• 1 <= capacity <= 3000
• 0 <= key <= \(10^4\)
• 0 <= value <= \(10^5\)
• At most 2 * \(10^5\) calls will be made to get and put.

Idea

Combine a hash map with a doubly linked list.

The linked list keeps the keys in order of use: the most recently used node sits right after a dummy head, and the least recently used node sits right before a dummy tail. The hash map maps each key to its list node, so any node can be found in \(O(1)\).

On get, look up the node and move it to the front. On put, update and move an existing node to the front, or insert a new node at the front; if the size then exceeds the capacity, remove the node before the tail and delete its key from the map. Because the list is doubly linked, removing or inserting a node only rewires its neighbors.

Time complexity: \(O(1)\) per get and put, Space complexity: \(O(capacity)\)

Pseudo Code

class Node {
    key, value, prev, next;
}

LRUCache(capacity) {
    this.capacity = capacity;
    map = new HashMap();    // key -> node
    head = new Node();      // dummy head, most recently used side
    tail = new Node();      // dummy tail, least recently used side
    head.next = tail;
    tail.prev = head;
}

get(key) {
    if (!map.contains(key)) {
        return -1;
    }
    node = map.get(key);
    remove(node);
    addToFront(node);
    return node.value;
}

put(key, value) {
    if (map.contains(key)) {
        node = map.get(key);
        node.value = value;
        remove(node);
        addToFront(node);
        return;
    }
    node = new Node(key, value);
    map.put(key, node);
    addToFront(node);
    if (map.size() > capacity) {
        lru = tail.prev;
        remove(lru);
        map.remove(lru.key);
    }
}

remove(node) {
    node.prev.next = node.next;
    node.next.prev = node.prev;
}

addToFront(node) {
    node.next = head.next;
    node.prev = head;
    head.next.prev = node;
    head.next = node;
}