https://leetcode.com/problems/lfu-cache/
很难，看了下面的参考：
https://discuss.leetcode.com/topic/69137/java-o-1-accept-solution-using-hashmap-doublelinkedlist-and-linkedhashset
注意其中的思想就是如下所述：
Your idea is brilliant... Especially storing all keys with same counts in one node,
if one of the keys in that node got hit once more,
it will be moved into a new node with (count+1) if the node exits or it will be wrapped into a newly created node with
(count+1). All your operations are guaranteed O(1) no doubt. There is no way to complete it bug-free within half an hour. So in the real interview,
I might as well explain the idea and how we should implement all operations in each scenario,
instead of actually trying to complete whole program...
Anyway, thank you so much for your time and explanation.
并且注意，用到了LinkedHashSet的特性，就是虽然是Set，但是是按照顺序插入的方式来遍历的。
public class LFUCache { private Node head = null; private int cap = 0; private HashMap<Integer, Integer> valueHash = null; private HashMap<Integer, Node> nodeHash = null; public LFUCache(int capacity) { this.cap = capacity; valueHash = new HashMap<Integer, Integer>(); nodeHash = new HashMap<Integer, Node>(); } public int get(int key) { if (valueHash.containsKey(key)) { increaseCount(key); return valueHash.get(key); } return -1; } public void set(int key, int value) { if ( cap == 0 ) return; if (valueHash.containsKey(key)) { valueHash.put(key, value); Node node = nodeHash.get(key); node.keys.remove(key); node.keys.add(key); } else { if (valueHash.size() < cap) { valueHash.put(key, value); } else { removeOld(); valueHash.put(key, value); } addToHead(key); } increaseCount(key); } private void addToHead(int key) { if (head == null) { head = new Node(0); head.keys.add(key); } else if (head.count > 0) { Node node = new Node(0); node.keys.add(key); node.next = head; head.prev = node; head = node; } else { head.keys.add(key); } nodeHash.put(key, head); } private void increaseCount(int key) { Node node = nodeHash.get(key); node.keys.remove(key); if (node.next == null) { node.next = new Node(node.count+1); node.next.prev = node; node.next.keys.add(key); } else if (node.next.count == node.count+1) { node.next.keys.add(key); } else { Node tmp = new Node(node.count+1); tmp.keys.add(key); tmp.prev = node; tmp.next = node.next; node.next.prev = tmp; node.next = tmp; } nodeHash.put(key, node.next); if (node.keys.size() == 0) remove(node); } private void removeOld() { if (head == null) return; int old = 0; for (int n: head.keys) { old = n; break; } head.keys.remove(old); if (head.keys.size() == 0) remove(head); nodeHash.remove(old); valueHash.remove(old); } private void remove(Node node) { if (node.prev == null) { head = node.next; } else { node.prev.next = node.next; } if (node.next != null) { node.next.prev = node.prev; } } class Node { public int count = 0; public LinkedHashSet<Integer> keys = null; public Node prev = null, next = null; public Node(int count) { this.count = count; keys = new LinkedHashSet<Integer>(); prev = next = null; } } } /** * Your LFUCache object will be instantiated and called as such: * LFUCache obj = new LFUCache(capacity); * int param_1 = obj.get(key); * obj.set(key,value); */