LFU算法

思路

LFU每个freq对应的集合其实就是个LRU，淘汰K时，会首先选取freq最小的集合，其次选取最久未使用的Key。
更新或者查询时，会将对应的K/V从当前freq提升至freq+1集合的末尾(假如头代表最久未使用)。

插入

查询

代码实现

···
class LFUCache {

private int minFreq;
 
private int capacity;
 
private Map<Integer, Integer> keyToVal;
 
private Map<Integer, Integer> keyToFreq;
 
private Map<Integer, LinkedHashSet<Integer>> freqToKeys;
 
public LFUCache(int capacity) {
    this.capacity = capacity;
    this.minFreq = 0;
    keyToVal = new HashMap<>();
    keyToFreq = new HashMap<>();
    freqToKeys = new HashMap<>();
}
 
public int get(int key) {
    if (!keyToVal.containsKey(key)) return -1;
    increaseFreq(key);
    return keyToVal.get(key);
}
 
public void put(int key, int value) {
    
    if (capacity <= 0) return;
 
    if (keyToVal.containsKey(key)) {
        keyToVal.put(key, value);
        increaseFreq(key);
        return;
    }
 
    if (keyToVal.size() == capacity) {
        removeMinKey();  
    }
 
    keyToVal.put(key, value);
    keyToFreq.put(key, 1);
    freqToKeys.compute(1, (k, v) -> v == null? new LinkedHashSet<>(): v).add(key);
    minFreq = 1;
}
 
private void increaseFreq(int key) {
    int freq = keyToFreq.get(key);
    freqToKeys.get(freq).remove(key);
    if (freqToKeys.get(freq).isEmpty()) {
        freqToKeys.remove(freq);
        if (freq == minFreq)
            minFreq++;
    }
    keyToFreq.put(key, freq + 1);
    freqToKeys.compute(freq + 1, (k, v) -> v == null? new LinkedHashSet<>(): v).add(key);
}
 
private void removeMinKey() {
    LinkedHashSet<Integer> set = freqToKeys.get(minFreq);
    int key = set.iterator().next();
    set.remove(key);
    if (set.isEmpty()) {
        freqToKeys.remove(minFreq);
    }
    keyToVal.remove(key);
    keyToFreq.remove(key);
}

}

/**

• Your LFUCache object will be instantiated and called as such:
• LFUCache obj = new LFUCache(capacity);
• int param_1 = obj.get(key);
• obj.put(key,value);
  */
  ···