package class19;

import java.util.HashMap;

// 本题测试链接 : https://leetcode.com/problems/lfu-cache/
// 提交时把类名和构造方法名改为 : LFUCache
public class Code02_LFUCache {

	public Code02_LFUCache(int K) {
		capacity = K;
		size = 0;
		records = new HashMap<>();
		heads = new HashMap<>();
		headList = null;
	}

	private int capacity; // 缓存的大小限制，即K
	private int size; // 缓存目前有多少个节点
	private HashMap<Integer, Node> records;// 表示key(Integer)由哪个节点(Node)代表
	private HashMap<Node, NodeList> heads; // 表示节点(Node)在哪个桶(NodeList)里
	private NodeList headList; // 整个结构中位于最左的桶

	// 节点的数据结构
	public static class Node {
		public Integer key;
		public Integer value;
		public Integer times; // 这个节点发生get或者set的次数总和
		public Node up; // 节点之间是双向链表所以有上一个节点
		public Node down;// 节点之间是双向链表所以有下一个节点

		public Node(int k, int v, int t) {
			key = k;
			value = v;
			times = t;
		}
	}

	// 桶结构
	public static class NodeList {
		public Node head; // 桶的头节点
		public Node tail; // 桶的尾节点
		public NodeList last; // 桶之间是双向链表所以有前一个桶
		public NodeList next; // 桶之间是双向链表所以有后一个桶

		public NodeList(Node node) {
			head = node;
			tail = node;
		}

		// 把一个新的节点加入这个桶，新的节点都放在顶端变成新的头部
		public void addNodeFromHead(Node newHead) {
			newHead.down = head;
			head.up = newHead;
			head = newHead;
		}

		// 判断这个桶是不是空的
		public boolean isEmpty() {
			return head == null;
		}

		// 删除node节点并保证node的上下环境重新连接
		public void deleteNode(Node node) {
			if (head == tail) {
				head = null;
				tail = null;
			} else {
				if (node == head) {
					head = node.down;
					head.up = null;
				} else if (node == tail) {
					tail = node.up;
					tail.down = null;
				} else {
					node.up.down = node.down;
					node.down.up = node.up;
				}
			}
			node.up = null;
			node.down = null;
		}
	}

	// removeNodeList：刚刚减少了一个节点的桶
	// 这个函数的功能是，判断刚刚减少了一个节点的桶是不是已经空了。
	// 1）如果不空，什么也不做
	//
	// 2)如果空了，removeNodeList还是整个缓存结构最左的桶(headList)。
	// 删掉这个桶的同时也要让最左的桶变成removeNodeList的下一个。
	//
	// 3)如果空了，removeNodeList不是整个缓存结构最左的桶(headList)。
	// 把这个桶删除，并保证上一个的桶和下一个桶之间还是双向链表的连接方式
	//
	// 函数的返回值表示刚刚减少了一个节点的桶是不是已经空了，空了返回true；不空返回false
	private boolean modifyHeadList(NodeList removeNodeList) {
		if (removeNodeList.isEmpty()) {
			if (headList == removeNodeList) {
				headList = removeNodeList.next;
				if (headList != null) {
					headList.last = null;
				}
			} else {
				removeNodeList.last.next = removeNodeList.next;
				if (removeNodeList.next != null) {
					removeNodeList.next.last = removeNodeList.last;
				}
			}
			return true;
		}
		return false;
	}

	// 函数的功能
	// node这个节点的次数+1了，这个节点原来在oldNodeList里。
	// 把node从oldNodeList删掉，然后放到次数+1的桶中
	// 整个过程既要保证桶之间仍然是双向链表，也要保证节点之间仍然是双向链表
	private void move(Node node, NodeList oldNodeList) {
		oldNodeList.deleteNode(node);
		// preList表示次数+1的桶的前一个桶是谁
		// 如果oldNodeList删掉node之后还有节点，oldNodeList就是次数+1的桶的前一个桶
		// 如果oldNodeList删掉node之后空了，oldNodeList是需要删除的，所以次数+1的桶的前一个桶，是oldNodeList的前一个
		NodeList preList = modifyHeadList(oldNodeList) ? oldNodeList.last : oldNodeList;
		// nextList表示次数+1的桶的后一个桶是谁
		NodeList nextList = oldNodeList.next;
		if (nextList == null) {
			NodeList newList = new NodeList(node);
			if (preList != null) {
				preList.next = newList;
			}
			newList.last = preList;
			if (headList == null) {
				headList = newList;
			}
			heads.put(node, newList);
		} else {
			if (nextList.head.times.equals(node.times)) {
				nextList.addNodeFromHead(node);
				heads.put(node, nextList);
			} else {
				NodeList newList = new NodeList(node);
				if (preList != null) {
					preList.next = newList;
				}
				newList.last = preList;
				newList.next = nextList;
				nextList.last = newList;
				if (headList == nextList) {
					headList = newList;
				}
				heads.put(node, newList);
			}
		}
	}

	public void put(int key, int value) {
		if (capacity == 0) {
			return;
		}
		if (records.containsKey(key)) {
			Node node = records.get(key);
			node.value = value;
			node.times++;
			NodeList curNodeList = heads.get(node);
			move(node, curNodeList);
		} else {
			if (size == capacity) {
				Node node = headList.tail;
				headList.deleteNode(node);
				modifyHeadList(headList);
				records.remove(node.key);
				heads.remove(node);
				size--;
			}
			Node node = new Node(key, value, 1);
			if (headList == null) {
				headList = new NodeList(node);
			} else {
				if (headList.head.times.equals(node.times)) {
					headList.addNodeFromHead(node);
				} else {
					NodeList newList = new NodeList(node);
					newList.next = headList;
					headList.last = newList;
					headList = newList;
				}
			}
			records.put(key, node);
			heads.put(node, headList);
			size++;
		}
	}

	public int get(int key) {
		if (!records.containsKey(key)) {
			return -1;
		}
		Node node = records.get(key);
		node.times++;
		NodeList curNodeList = heads.get(node);
		move(node, curNodeList);
		return node.value;
	}

}