题解 | #设计LRU缓存结构#

public class Solution {
    /**
     * lru design
     * @param operators int整型二维数组 the ops
     * @param k int整型 the k
     * @return int整型一维数组
     */
    public int[] LRU (int[][] operators, int k){
        // write code here
        if( k <= 0 || operators == null){
            return null;
        }
        int resultLength = 0;
        for(int i = 0;i< operators.length;i++){
            if(operators[i][0] == 2){
                resultLength ++;
            }
        }
        int index = 0;
        int[] result = new int[resultLength];
        MyLruCache<Integer,Integer> cache = new MyLruCache<>(k);
        for(int i = 0;i<operators.length;i++){
            switch(operators[i][0]){
                case 1:
                    cache.set(operators[i][1],operators[i][2]);
                    break;
                case 2:
                    result[index ++] = cache.get(operators[i][1]) == null ? -1 : cache.get(operators[i][1]);
                    break;
                default:
                    break;
            }
        }
        return result;
    }
    
    public class MyLruCache<K,V>{
        private int cacheSize;
        private float factory = 0.75f;
        private LinkedHashMap<K,V> map;
        public MyLruCache(int cacheSize) {
            this.cacheSize = cacheSize;
            map = new LinkedHashMap<K,V>(cacheSize, factory, true){
                protected boolean removeEldestEntry(Map.Entry<K, V> eldest){
                    return size() > MyLruCache.this.cacheSize;
                };
            };
        }
        public V get(K key){

            return map.get(key);

        }
        public void set(K key,V value){
            map.put(key,value);
        }
    }
    
    
    
    
    
    
}