LRU Cache
Similar Problems:
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.
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
Github: code.dennyzhang.com
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/lru-cache ## Basic Ideas: Double linked list + hashmap ## ## We need to mapping from dlink to hashmap, ## and also from hashmap to dlink ## ## Complexity: class DLinkNode: def __init__(self, val): self.val = val self.left, self.right = None, None class LRUCache: def __init__(self, capacity): """ :type capacity: int """ self.count, self.capacity = 0, capacity self.d = {} # double link. head is the delete candidates self.head, self.tail = DLinkNode(None), DLinkNode(None) self.head.right = self.tail self.tail.left = self.head def get(self, key): """ :type key: int :rtype: int """ if key in self.d: p = self.d[key] # remove the node p.left.right = p.right p.right.left = p.left # add to the tail p.left = self.tail.left p.left.right = p p.right = self.tail p.right.left = p return p.val[1] else: return -1 def put(self, key, value): """ :type key: int :type value: int :rtype: void """ if key in self.d: self.get(key) self.tail.left.val = (key, value) return # need to add one more node if self.count == self.capacity: # remove one from the head p = self.head.right p.left.right = p.right p.right.left = p.left # update related info self.count -= 1 del self.d[p.val[0]] # create a new node p = DLinkNode((key, value)) # add to tail p.left = self.tail.left p.left.right = p p.right = self.tail p.right.left = p # update related info self.count += 1 self.d[key] = p # 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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