Description:
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
Solutions:
Brute force:
from collections import deque
class LRUCache:
def __init__(self, capacity: int):
self.q = deque()
self.dic = {}
self.cap = capacity
self.count = 0
def get(self, key: int) -> int:
if key in self.dic:
self.q.remove(key)
self.q.append(key)
return self.dic[key]
return -1
def put(self, key: int, value: int) -> None:
self.count += 1
if key not in self.dic:
self.q.append(key)
else:
self.q.remove(key)
self.q.append(key)
self.dic[key] = value
if len(self.q) > self.cap:
self.dic.pop(self.q.popleft())
Hashtable + double linked list: 野路子
class Node:
def __init__(self,prev,key,value,next_,):
self.prev = prev
self.key = key
self.val = value
self.next = next_
class LRUCache:
def __init__(self, capacity: int):
self.cap = capacity+1
self.zero = Node(None,None,None,None)
self.left = self.zero
self.right = self.zero
self.dic = {None:self.zero}
self.count = 1
def get(self, key: int) -> int:
if key in self.dic:
if self.dic[key] != self.right:
prev = self.dic[key].prev
next_ = self.dic[key].next
prev.next = next_
next_.prev = prev
if self.dic[key] == self.left:
self.left = self.zero.next
self.dic[key].prev = self.right
self.right.next = self.dic[key]
self.right = self.dic[key]
return self.dic[key].val
return -1
def put(self, key: int, value: int) -> None:
if key not in self.dic:
self.count += 1
self.dic[key] = Node(self.right,key,value,None)
if self.count == 2:
self.left = self.dic[key]
self.right.next = self.dic[key]
self.right = self.dic[key]
if self.count > self.cap:
self.dic.pop(self.left.key)
self.left = self.left.next
self.zero.next = self.left
self.left.prev = self.zero
self.count -= 1
elif self.dic[key] == self.right:
self.dic[key].val = value
else:
prev = self.dic[key].prev
next_ = self.dic[key].next
prev.next = next_
next_.prev = prev
if self.dic[key] == self.left:
self.left = self.zero.next
self.dic[key].val = value
self.dic[key].prev = self.right
self.right.next = self.dic[key]
self.right = self.dic[key]
Runtime: 244 ms, faster than 43.29% of Python3 online submissions for LRU Cache.
Memory Usage: 23.2 MB, less than 5.07% of Python3 online submissions for LRU Cache.
sample 192 ms submission: Cheating solution
NOTE: 重点在于使用del self.d[next(iter(self.d))]删除头部元素。看来是Python dictionary内置的功能了。
class LRUCache:
def __init__(self, capacity: int):
self.cap = capacity
self.d = {}
def get(self, key: int) -> int:
if key in self.d:
value = self.d[key]
else:
return -1
# move the key to the 'end' of the dictionary
del self.d[key]
self.d[key] = value
return value
def put(self, key: int, value: int) -> None:
# place the key at the 'end' of the dictionary
if key in self.d:
del self.d[key]
self.d[key] = value
# if over capacity, delete the 'front' of the dictionary
if len(self.d) > self.cap:
del self.d[next(iter(self.d))]
Using OrderedDict
from collections import OrderedDict
class LRUCache:
def __init__(self, capacity: int):
self.cap = capacity
self.dic = OrderedDict()
def get(self, key: int) -> int:
if key in self.dic:
self.dic.move_to_end(key)
return self.dic[key]
return -1
def put(self, key: int, value: int) -> None:
if key not in self.dic:
self.dic[key] = value
if len(self.dic) > self.cap:
self.dic.popitem(last=False)
else:
self.dic[key] = value
self.dic.move_to_end(key)
Runtime: 192 ms, faster than 99.80% of Python3 online submissions for LRU Cache.
Memory Usage: 23 MB, less than 5.07% of Python3 online submissions for LRU Cache.
CPP Solution: combine linked_list and hashmap
Inspired by https://zxi.mytechroad.com/blog/hashtable/leetcode-146-lru-cache/
class LRUCache {
public:
LRUCache(int capacity_) {
capacity = capacity_;
}
int get(int key) {
const auto it = dic.find(key);
if (it != dic.cend()){
ls.splice(ls.begin(),ls,it->second);
return it->second->second;
}
return -1;
}
void put(int key, int value) {
const auto it = dic.find(key);
if (it != dic.cend()){
ls.splice(ls.begin(),ls,it->second);
it->second->second = value;
}
else{
ls.emplace_front(key,value);
dic[key] = ls.begin();
if (ls.size() > capacity){
dic.erase(ls.back().first);
ls.pop_back();
}
}
}
private:
int capacity;
list> ls;
unordered_map>::iterator> dic;
};
Runtime: 88 ms, faster than 99.62% of C++ online submissions for LRU Cache.
Memory Usage: 38 MB, less than 78.53% of C++ online submissions for LRU Cache.