8. LRU Cache

[bonfy]

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

Leetcode: 146. LRU Cache - https://leetcode.com/problems/lru-cache

[bonfy]

解法：

注意： Get 和 Put 之后 key 都会成为 Recent 那个

class LRUCache:

    def __init__(self, capacity):
        self.dic = collections.OrderedDict()
        self.remain = capacity

    def get(self, key):
        if key not in self.dic:
            return -1
        self.dic.move_to_end(key)
        return self.dic[key]

    def put(self, key, value):
        if key in self.dic:    
            self.dic.move_to_end(key)
        else:
            if self.remain > 0:
                self.remain -= 1  
            else:  # self.dic is full
                self.dic.popitem(last=False) 
        self.dic[key] = value