[leetcode] 146. LRU Cache

문제 설명

링크
LRU 캐시 구현

접근 - Doubly Linked List, Dict

• dict[key] = node
• head, tail
• head : 가장 최근 사용
• tail : lru

코드

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


class LRUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.size = 0
        self.table = {}
        self.head = None
        self.tail = None

    def get(self, key: int) -> int:
        if self.table.get(key):
            node = self.table[key]
            self.removeItem(node)
            self.insertHead(node)
            return node.val
        else:
            return -1

    def put(self, key: int, value: int) -> None:
        if self.table.get(key):
            node = self.table[key]
            self.removeItem(node)
            self.insertHead(node)
            node.val = value
            return
        node = Node(key, value)

        if self.size >= self.capacity:
            self.removeTail()
        else:
            self.size += 1
        self.insertHead(node)

    def removeTail(self):
        if len(self.table) == 0: return
        del self.table[self.tail.key]
        if self.head == self.tail:
            self.head = self.tail = None
        else:
            self.tail = self.tail.prev
            self.tail.next = None

    def insertHead(self, node: 'Node'):
        node.next = self.head

        if self.head:
            self.head.prev = node

        self.head = node
        node.prev = None

        if not self.tail:
            self.tail = node
        self.table[node.key] = node

    def removeItem(self, node: 'Node'):
        prev = node.prev
        next = node.next
        if prev:
            prev.next = next
        else:
            self.head = next
        if next:
            next.prev = prev
        else:
            self.tail = prev