​LeetCode刷题实战146：LRU 缓存机制

输入
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
输出
[null, null, null, 1, null, -1, null, -1, 3, 4]

解释
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // 缓存是 {1=1}
lRUCache.put(2, 2); // 缓存是 {1=1, 2=2}
lRUCache.get(1); // 返回 1
lRUCache.put(3, 3); // 该操作会使得关键字 2 作废，缓存是 {1=1, 3=3}
lRUCache.get(2); // 返回 -1 (未找到)
lRUCache.put(4, 4); // 该操作会使得关键字 1 作废，缓存是 {4=4, 3=3}
lRUCache.get(1); // 返回 -1 (未找到)
lRUCache.get(3); // 返回 3
lRUCache.get(4); // 返回 4

/**
 * HashMap.Node subclass for normal LinkedHashMap entries.
 */
static class Entry extends HashMap.Node {
     Entry before, after;
     Entry(int hash, K key, V value, Node next) {
          super(hash, key, value, next);
     }
}

class LRUCache {
 
    private int capacity;
    private LRULinkedHashMap lruLinkedHashMap = new LRULinkedHashMap<>();
 
    private class LRULinkedHashMap<K, V> extends LinkedHashMap<K, V> {
        @Override
        protected boolean removeEldestEntry(Map.Entry eldest) {
            if (size() > capacity) {
                return true;
            } else {
                return false;
            }
        }
    }
 
    public LRUCache(int capacity) {
        this.capacity = capacity;
    }
 
    public int get(int key) {
        Integer value = lruLinkedHashMap.get(key);
        if (null == value) {
            return -1;
        }
        lruLinkedHashMap.remove(key);
        lruLinkedHashMap.put(key, value);
        return value;
    }
 
    public void put(int key, int value) {
        if (lruLinkedHashMap.containsKey(key)) {
            lruLinkedHashMap.remove(key);
        }
        lruLinkedHashMap.put(key, value);
    }
 
}