Leetcode 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.

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 1 1 <= capacity <= 3000 3000 3000
  • 0 0 0 <= key <= 1 0 4 10^4 104
  • 0 0 0 <= value <= 1 0 5 10^5 105
  • At most 2 ∗ 1 0 5 2 * 10^5 2105 calls will be made to get and put.

Algorithm

map and doubly linked list.
https://www.code-recipe.com/post/lru-cache

Code

class LRUCache:
    def __init__(self, capacity: int):
        class node:
            def __init__(self, key, val, L=None, R=None):
                self.key = key
                self.val = val
                self.L = L
                self.R = R
        
        class dlink:
            def __init__(self):
                self.head = node(-1, -1)
                self.tail = node(-1, -1)
                self.head.R = self.tail
                self.tail.L = self.head
                self.len = 0

            def add_node(self, key, val):
                new_node = node(key, val, self.head, self.tail)
                new_node.L = self.head
                new_node.R = self.head.R
                self.head.R.L = new_node
                self.head.R = new_node
                self.len += 1
                return new_node

            def del_node(self, now_node=None):
                if not now_node:
                    now_node = self.tail.L
                now_node.L.R = now_node.R
                now_node.R.L = now_node.L
                self.len -= 1
                return now_node

        self.len = capacity
        self.map = {}
        self.lik = dlink()

    def get(self, key: int) -> int:
        if key in self.map:
            index = self.lik.del_node(self.map[key])
            del self.map[index.key]
            new_node = self.lik.add_node(index.key, index.val)
            self.map[key] = new_node
            return new_node.val
        return -1

    def put(self, key: int, value: int) -> None:
        if key in self.map:
            self.lik.del_node(self.map[key])
            del self.map[key]
        
        new_node = self.lik.add_node(key, value)
        self.map[key] = new_node 
        
        if len(self.map) > self.len:
            now_node = self.lik.del_node() # Remove last node
            del self.map[now_node.key]

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

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