leetcode146 LRU Cache

"""
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
"""
"""
用dict存储元素
用list存储dict里的key
用list里的insert函数和index函数pop函数来维护队列
保证能在队头加元素，删除任意index位置元素
"""
class LRUCache:

    def __init__(self, capacity: int):
        self.len = capacity
        self.d = {}
        self.l = []

    def get(self, key: int) -> int:
        val = self.d.get(key)
        if val and self.l[0] != key: #如果查找的值不在队头
            index = self.l.index(key)
            self.l.pop(index)
            self.l.insert(0, key)
        val = val if val else -1
        return val

    def put(self, key: int, value: int) -> None:
        if self.d.get(key): #如果重复
            index = self.l.index(key)
            self.d.pop(key)
            self.l.pop(index)
        if len(self.l) >= self.len: #如果队满
            x = self.l.pop(-1)
            self.d.pop(x)
        self.d[key] = value
        self.l.insert(0, key)