146. LRU Cache #

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.

Follow up:Could you do get and put in O(1) 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 <= 3000
0 <= value <= 104
At most 3 * 104 calls will be made to get and put.

Problem Summary #

Use the data structures you have mastered to design and implement an LRU (Least Recently Used) cache mechanism. Implement the LRUCache class:

LRUCache(int capacity) Initialize the LRU cache with a positive integer capacity
int get(int key) If the keyword key exists in the cache, return the value of the keyword; otherwise return -1.
void put(int key, int value) If the keyword already exists, change its data value; if the keyword does not exist, insert this “keyword-value” pair. When the cache capacity reaches its upper limit, it should delete the least recently used data value before writing new data, thereby making room for the new data value.

Follow-up: Can you complete both operations in O(1) time complexity?

Solution Ideas #

This problem is a classic LRU interview question. See the Chapter 3 template for a detailed explanation.

Code #

package leetcode

type LRUCache struct {
	head, tail *Node
	Keys       map[int]*Node
	Cap        int
}

type Node struct {
	Key, Val   int
	Prev, Next *Node
}

func Constructor(capacity int) LRUCache {
	return LRUCache{Keys: make(map[int]*Node), Cap: capacity}
}

func (this *LRUCache) Get(key int) int {
	if node, ok := this.Keys[key]; ok {
		this.Remove(node)
		this.Add(node)
		return node.Val
	}
	return -1
}

func (this *LRUCache) Put(key int, value int) {
	if node, ok := this.Keys[key]; ok {
		node.Val = value
		this.Remove(node)
		this.Add(node)
		return
	} else {
		node = &Node{Key: key, Val: value}
		this.Keys[key] = node
		this.Add(node)
	}
	if len(this.Keys) > this.Cap {
		delete(this.Keys, this.tail.Key)
		this.Remove(this.tail)
	}
}

func (this *LRUCache) Add(node *Node) {
	node.Prev = nil
	node.Next = this.head
	if this.head != nil {
		this.head.Prev = node
	}
	this.head = node
	if this.tail == nil {
		this.tail = node
		this.tail.Next = nil
	}
}

func (this *LRUCache) Remove(node *Node) {
	if node == this.head {
		this.head = node.Next
		node.Next = nil
		return
	}
	if node == this.tail {
		this.tail = node.Prev
		node.Prev.Next = nil
		node.Prev = nil
		return
	}
	node.Prev.Next = node.Next
	node.Next.Prev = node.Prev
}