刚刚发现，老师录制视频的时候，这道题还是 Hard ，结果现在就已经变成 Medium 了。

茶艺师学编程 超哥板书 146. LRU 缓存机制 class LRUCache: def __init__(self, capacity: int): self.dic = collections.OrderedDict() self.remain = capacity def get(self, key: int) -> int: if key not in self.dic: return -1 v = self.dic.pop(key) self.dic[key] = v return v def put(self, key: int, value: int) -> None: if key in self.dic: self.dic.pop(key) else: if self.remain > 0: self.remain -= 1 else: self.dic.popitem(last=False) self.dic[key] = value

class LRUCache { private LinkedHashMap<Integer, Integer> cache; private int size; public LRUCache(int capacity) { cache = new LinkedHashMap<Integer, Integer>(); size = capacity; } public int get(int key) { if (!cache.containsKey(key)) { return -1; } int value = cache.remove(key); cache.put(key, value); return value; } public void put(int key, int value) { if (cache.containsKey(key)) { cache.remove(key); } else { if (size > 0) { size--; } else { // 移除第一个原始 cache.remove(cache.entrySet().iterator().next().getKey()); } } cache.put(key, value); } public void print() { Set<Entry<Integer, Integer>> keys = cache.entrySet(); System.out.print("[ "); for (Entry<Integer, Integer> entry : keys) { System.out.print(entry.getKey() + ":" + entry.getValue() + ", "); } System.out.println(" ]"); } }