力扣labuladong——一刷day89

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文章目录

• 前言
• 一、力扣460. LFU 缓存

前言

LFU 算法是要复杂很多的，而且经常出现在面试中，因为 LFU 缓存淘汰算法在工程实践中经常使用，也有可能是因为 LRU 算法太简单了。不过话说回来，这种著名的算法的套路都是固定的，关键是由于逻辑较复杂，不容易写出漂亮且没有 bug 的代码

一、力扣460. LFU 缓存

class LFUCache {private int cap;private int minFreq;private HashMap<Integer,Integer> keyToVal;private HashMap<Integer,Integer> keyToFreq;private HashMap<Integer,LinkedHashSet<Integer>> freqToKeys;public LFUCache(int capacity) {this.cap = capacity;minFreq = 0;keyToVal = new HashMap<>();keyToFreq = new HashMap<>();freqToKeys = new HashMap<>();}public int get(int key) {if(!keyToVal.containsKey(key)){return -1;}increase(key);return keyToVal.get(key);}public void put(int key, int value) {if(cap <= 0)return;if(keyToVal.containsKey(key)){keyToVal.put(key,value);increase(key);return;}if(cap <= keyToVal.size()){removeMinFreq();}keyToVal.put(key,value);keyToFreq.put(key,1);freqToKeys.putIfAbsent(1,new LinkedHashSet<Integer>());freqToKeys.get(1).add(key);this.minFreq = 1;}private void increase(int key){int freq = keyToFreq.get(key);keyToFreq.put(key,freq+1);freqToKeys.get(freq).remove(key);freqToKeys.putIfAbsent(freq+1,new LinkedHashSet<Integer>());freqToKeys.get(freq+1).add(key);if(freqToKeys.get(freq).isEmpty()){freqToKeys.remove(freq);if(freq == minFreq){this.minFreq ++;}}}private void removeMinFreq(){LinkedHashSet<Integer> list = freqToKeys.get(this.minFreq);int key = list.iterator().next();list.remove(key);if(list.isEmpty()){freqToKeys.remove(this.minFreq);}keyToFreq.remove(key);keyToVal.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); */