# Question

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.

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.put(4, 4);    // evicts key 1
cache.get(3);       // returns 3
cache.get(4);       // returns 4


# Solution

Use doubly Linked List to track access history, and use dict to track key-value mapping. The value in dict should be node in the doubly linked list.

class Node:
def __init__(self, k, v):
self.key = k
self.val = v
self.prev = None
self.next = None

class LRUCache(object):

def __init__(self, capacity):
"""
:type capacity: int
"""
self.capacity = capacity
self.dic = dict()

self.tail = Node(0, 0)

def get(self, key):
"""
:type key: int
:rtype: int
"""
if key in self.dic:
n = self.dic[key]
self._remove(n)
return n.val
return -1

def put(self, key, value):
"""
:type key: int
:type value: int
:rtype: void
"""
if key in self.dic:
self._remove(self.dic[key])
n = Node(key, value)
self.dic[key] = n
if len(self.dic) > self.capacity:
self._remove(n)
del self.dic[n.key]

def _remove(self, node):
"""
Remove the node from the doubly-linked-list
"""
p = node.prev
n = node.next
p.next = n
n.prev = p

"""
Insert the node to tail
"""
p = self.tail.prev
p.next = node
self.tail.prev = node
node.prev = p
node.next = self.tail

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