LeetCode 146. LRU Cache

Description

https://leetcode.com/problems/lru-cache/

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.

The cache is initialized with a positive capacity.

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

Explanation

use linked list and hashmap together

Python Solution

class ListNode:
    def __init__(self, key, value):
        self.key = key
        self.value = value
        self.next = None
        self.prev = None
        

class LRUCache:        
    def __init__(self, capacity: int):
        self.cache = {}
        self.count = 0
        self.capacity = capacity
        
        self.head = ListNode(-1, -1)
        self.tail = ListNode(-1, -1)
        self.head.next = self.tail
        self.tail.prev = self.head

    def get(self, key: int) -> int:
        
        node = self.cache.get(key, None)
        
        if not node:        
            return -1
        
        self.insert_to_head(node)
        
        return node.value
        
    def put(self, key: int, value: int) -> None:
        node = self.cache.get(key)
        
        if not node:
            new_node = ListNode(key, value)
            
            self.cache[key] = new_node
            self.add_node(new_node)
                
            self.count += 1
        
            if self.count > self.capacity:   
                tail = self.remove_tail()
                del self.cache[tail.key]
                self.count -= 1
        else:
            node.value = value
            self.insert_to_head(node)
        
    def add_node(self, node):
        node.prev = self.head
        node.next = self.head.next
        
        self.head.next.prev = node
        self.head.next = node
        
    def remove_node(self, node):
        prev = node.prev
        new = node.next
        
        prev.next = new
        new.prev = prev       
        
    def insert_to_head(self, node):
        self.remove_node(node)
        self.add_node(node)        
        
    def remove_tail(self):
        res = self.tail.prev
        self.remove_node(res)
        return res          
# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)
  • Time complexity: O(N).
  • Space complexity: O(N).

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