Leetcode 146. LRU cache

Title Description

Please design and implement a meeting LRU ( Recently at least use ) cache Constrained data structure .
Realization LRUCache class ：
LRUCache(int capacity) With Positive integer As capacity capacity initialization LRU cache
int get(int key) If the keyword key In the cache , The value of the keyword is returned , Otherwise return to -1 .
void put(int key, int value) If the keyword key Already exist , Change its data value value ; If it doesn't exist , Then insert the group into the cache key-value . If the insert operation causes the number of keywords to exceed capacity , Should be Eviction The longest unused keyword .
function get and put Must be O(1) The average time complexity of running .

 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]

 explain 
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); //  The cache is  {
    1=1}
lRUCache.put(2, 2); //  The cache is  {
    1=1, 2=2}
lRUCache.get(1);    //  return  1
lRUCache.put(3, 3); //  This operation causes the keyword to  2  To void , The cache is  {
    1=1, 3=3}
lRUCache.get(2);    //  return  -1 ( Not found )
lRUCache.put(4, 4); //  This operation causes the keyword to  1  To void , The cache is  {
    4=4, 3=3}
lRUCache.get(1);    //  return  -1 ( Not found )
lRUCache.get(3);    //  return  3

Solution 1 ： Hashtable + Linked list

This problem is the same as that of data flow , It's all a routine ： Use a linked list to record the data sequence , Use hash table to quickly find linked list nodes , Then move the node location according to the demand . Here we use a one-way linked list to realize , Hash table storage key Corresponding value The previous node of the node . Of course, you can also directly use a two-way linked list .

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


class LRUCache:
    """ solution 1：HashMap + LinkedList"""
    def __init__(self, capacity: int):
        self.dummy = ListNode(-1, -1)
        self.tail = self.dummy
        self.key_to_prev = {
    }
        self.capacity = capacity

    def get(self, key: int) -> int:
        if key not in self.key_to_prev:
            return -1
        self.remove_to_last(key)
        return self.key_to_prev.get(key).next.value

    def put(self, key: int, value: int) -> None:
        if key in self.key_to_prev:
            self.remove_to_last(key)
            self.tail.value = value
            return
        if len(self.key_to_prev) < self.capacity:
            self.add_key_value(key, value)
            return
            # not enough capacity
        self.remove_lru()
        self.add_key_value(key, value)

    def set_value(self, key, value):
        pre_node = self.key_to_prev.get(key)
        target_node = pre_node.next
        target_node.value = value

    def add_key_value(self, key, value):
        new_node = ListNode(key, value)
        self.key_to_prev[key] = self.tail  # s
        self.tail.next = new_node
        self.tail = new_node

    def remove_lru(self):
        remove_target = self.dummy.next
        self.key_to_prev.pop(remove_target.key)
        self.dummy.next = remove_target.next
        # capacity  May be 1, So we still have to judge target.next Is it empty 
        if remove_target == self.tail:
            self.tail = self.dummy
        else:
            self.key_to_prev[remove_target.next.key] = self.dummy

    def remove_to_last(self, key):
        pre = self.key_to_prev[key]
        target = pre.next
        if self.tail == target:
            return
        pre.next = target.next

        self.key_to_prev[target.next.key] = pre
        target.next = None
        self.tail.next = target
        self.key_to_prev[key] = self.tail
        self.tail = target

Solution 2 ：OrderedDict

Directly use the built-in data structure ：OrderedDict.

class LRUCache:
    """  solution 2：OrderedDict """
    def __init__(self, capacity: int):
        self.cache = OrderedDict()
        self.capacity = capacity

    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        value = self.cache.pop(key)  #  Visited ,  Delete first 
        self.cache[key] = value  #  Add to the end 
        return value

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            self.cache.pop(key)  #  Delete first 
        self.cache[key] = value  #  Add to the end 
        if len(self.cache) > self.capacity:
            self.cache.popitem(last=False)