Design and implement a data structure for a Least Frequently Used (LFU) cache.
Implement the LFUCache class:
LFUCache(int capacity)Initializes the object with thecapacityof the data structure.int get(int key)Gets the value of thekeyif thekeyexists in the cache. Otherwise, returns-1.void put(int key, int value)Update the value of thekeyif present, or inserts thekeyif not already present. When the cache reaches itscapacity, it should invalidate and remove the least frequently used key before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently usedkeywould be invalidated.
To determine the least frequently used key, a use counter is maintained for each key in the cache. The key with the smallest use counter is the least frequently used key.
When a key is first inserted into the cache, its use counter is set to 1 (due to the put operation). The use counter for a key in the cache is incremented either a get or put operation is called on it.
The functions get and put must each run in O(1) average time complexity.
Example 1:
Input ["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"] [[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]] Output [null, null, null, 1, null, -1, 3, null, -1, 3, 4] Explanation // cnt(x) = the use counter for key x // cache=[] will show the last used order for tiebreakers (leftmost element is most recent) LFUCache lfu = new LFUCache(2); lfu.put(1, 1); // cache=[1,_], cnt(1)=1 lfu.put(2, 2); // cache=[2,1], cnt(2)=1, cnt(1)=1 lfu.get(1); // return 1 // cache=[1,2], cnt(2)=1, cnt(1)=2 lfu.put(3, 3); // 2 is the LFU key because cnt(2)=1 is the smallest, invalidate 2. // cache=[3,1], cnt(3)=1, cnt(1)=2 lfu.get(2); // return -1 (not found) lfu.get(3); // return 3 // cache=[3,1], cnt(3)=2, cnt(1)=2 lfu.put(4, 4); // Both 1 and 3 have the same cnt, but 1 is LRU, invalidate 1. // cache=[4,3], cnt(4)=1, cnt(3)=2 lfu.get(1); // return -1 (not found) lfu.get(3); // return 3 // cache=[3,4], cnt(4)=1, cnt(3)=3 lfu.get(4); // return 4 // cache=[3,4], cnt(4)=2, cnt(3)=3
参考：https://leetcode.com/problems/lfu-cache/discuss/94521/JAVA-O(1)-very-easy-solution-using-3-HashMaps-and-LinkedHashSet
没啥好说的，直接抄吧
public class LFUCache {
private int min;
private final int capacity;
//keyToVal存储每次put的value
private final HashMap<Integer, Integer> keyToVal;
//keyToCount存储访问次数
private final HashMap<Integer, Integer> keyToCount;
private final HashMap<Integer, LinkedHashSet<Integer>> countToLRUKeys;
public LFUCache(int capacity) {
this.min = -1;
this.capacity = capacity;
this.keyToVal = new HashMap<>();
this.keyToCount = new HashMap<>();
this.countToLRUKeys = new HashMap<>();
}
public int get(int key) {
if (!keyToVal.containsKey(key)) return -1;
int count = keyToCount.get(key);
countToLRUKeys.get(count).remove(key); //从当前计数中删除键（因为我们将重新计算计数）remove key from current count (since we will inc count)
if (count == min && countToLRUKeys.get(count).size() == 0) min++; // nothing in the current min bucket 当前最小存储桶中没有任何内容的时候，min++
//添加和返回个数，每get+1，访问次数+1
putCount(key, count + 1);
return keyToVal.get(key);
}
public void put(int key, int value) {
if (capacity <= 0) return;
if (keyToVal.containsKey(key)) {
keyToVal.put(key, value); //更新key update key‘s value
get(key); //更新密钥的count update key‘s count
return;
}
if (keyToVal.size() >= capacity)
evict(countToLRUKeys.get(min).iterator().next()); // evict LRU from this min count bucket 从这个最小计数桶中驱逐最不常读取的元素LRU
min = 1; //min 重新计算为1
putCount(key, min); //adding new key and count 添加新键和计数
keyToVal.put(key, value); //adding new key and value 添加新的键和值
}
//从两个map中删掉
private void evict(int key) {
countToLRUKeys.get(min).remove(key);
keyToVal.remove(key);
}
//给两个map添加count
private void putCount(int key, int count) {
keyToCount.put(key, count);
countToLRUKeys.computeIfAbsent(count, ignore -> new LinkedHashSet<>());
countToLRUKeys.get(count).add(key);
}
}