Leetcode 146 - LRU cache

Note:

  • To get something is O(1), that must be a hash.
  • But hash has no order. How can we know the updating order? We need another data structure.
  • It can’t be an array because searching is at least O(n) for array. It can’t be a single linked list coz seaching is also O(n).
  • Let’s use a double linked list! And store the reference in hash.
  • Use dummyHead and dummyTail are techniques in linked list to bring convenience.
  • We need several helpers
    • moveToHead(node).
    • addToHead(node).
    • removeLRUItem()
    • removeNodeFromList(node).

Question:

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:

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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

Code:

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/**
* @param {number} capacity
*/
var LRUCache = function(capacity) {
this.capacity = capacity;
this.hash = {};
this.count = 0;
this.dummyHead = new ListNode();
this.dummyTail = new ListNode();
this.dummyHead.next = this.dummyTail;
this.dummyTail.prev = this.dummyHead;
};

/**
* @param {number} key
* @return {number}
*/
LRUCache.prototype.get = function(key) {
let node = this.hash[key];
if (!node) return -1;
this.moveToHead(node);
return node.val;
};

/**
* @param {number} key
* @param {number} value
* @return {void}
*/
LRUCache.prototype.put = function(key, value) {
if (this.hash[key]) {
let node = this.hash[key];
node.val = value;
this.moveToHead(node);
} else {
let node = new ListNode(key, value);
if (this.count === this.capacity) {
this.removeLRUItem();
}
this.addToHead(node);
this.hash[key] = node;
this.count++;
}
};

LRUCache.prototype.moveToHead = function(node) {
// Remove it
this.removeNodeFromList(node);
// Add
this.addToHead(node);
}

LRUCache.prototype.removeNodeFromList = function(node) {
node.prev.next = node.next;
node.next.prev = node.prev;
}

LRUCache.prototype.removeLRUItem = function() {
const tail = this.dummyTail.prev;
this.removeNodeFromList(tail);
delete this.hash[tail.key];
this.count--;
}

LRUCache.prototype.addToHead = function(node) {
// Add to head;
node.prev = this.dummyHead;
node.next = this.dummyHead.next;
this.dummyHead.next.prev = node;
this.dummyHead.next = node;
}

var ListNode = function (key, val) {
this.key = key;
this.val = val;
this.next = null;
this.prev = null;
}

/**
* Your LRUCache object will be instantiated and called as such:
* var obj = new LRUCache(capacity)
* var param_1 = obj.get(key)
* obj.put(key,value)
*/