146. LRU Cache

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

The cache is initialized with a positive capacity.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

class LRUCache {
    private Map<Integer, Integer> map;
    private int capacity;
    public LRUCache(int capacity) {
        this.capacity = capacity;
        map = new LinkedHashMap<Integer, Integer>();
    }
    
    public int get(int key) {
        Integer val = map.get(key);
        if (val == null) return -1;
        map.remove(key);
        map.put(key, val);//确保get后当前key变成最后一个插入的
        return val;
    }
    
    public void put(int key, int value) {
        map.remove(key);//Why? Because假如update键的新value也算作visited，要先把当前key删掉重新insert确保是最后一个插入的
        map.put(key, value);
        if (map.size() > capacity){
            // for(Map.Entry<Integer, Integer> entry: map.entrySet()){
            //     System.out.println("key"+entry.getKey()+"value"+entry.getValue());
            // }
            map.remove(map.entrySet().iterator().next().getKey());
        }
    }
}

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

来自小土刀大神，短小精悍，用了LinkedHashMap的特性（保留插入顺序）。

get的时候先取出value，然后重新插入确保这个key是最后一个值。

put的时候也是要先remove了再重新put进去保证update键的情况

class LRUCache {
  
  Node head = new Node(0, 0), tail = new Node(0, 0);
  Map<Integer, Node> map = new HashMap();
  int capacity;
  
  public LRUCache(int _capacity) {
    capacity = _capacity;
    head.next = tail;
    tail.prev = head;
  }

  public int get(int key) {
    if(map.containsKey(key)) {
      Node node = map.get(key);
      remove(node);
      insert(node);
      return node.value;
    } else {
      return -1;
    }
  }

  public void put(int key, int value) {
    if(map.containsKey(key)) {
      remove(map.get(key));
    }
    if(map.size() == capacity) {
      remove(tail.prev);
    }
    insert(new Node(key, value));
  }
  
  private void remove(Node node) {
    map.remove(node.key);
    node.prev.next = node.next;
    node.next.prev = node.prev;
  }
  
  private void insert(Node node){
    map.put(node.key, node);
    Node headNext = head.next;
    head.next = node;
    node.prev = head;
    headNext.prev = node;
    node.next = headNext;
  }
  
  class Node{
    Node prev, next;
    int key, value;
    Node(int _key, int _value) {
      key = _key;
      value = _value;
    }
  }
}

用linkedhashmap太赖了，用一般hashmap估计能报名

原理：Node：prev，next，key，val

然后用一个头节点head，尾节点tail执行insert和remove操作，insert每次加到最前边，也就是head。next。remove每次把tail.prev移除掉，注意更新节点的prev,next