使用LinkedHashMap实现一个简易的LRU缓存

Posted 2021-02-21 Rainyn

leetcode 解法

https://blog.csdn.net/Apeopl/article/details/90137398

import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Set;

public class SimpleLRUCache<K, V> {

    private final int MAX_CACHE_SIZE;
    private final float DEFAULT_LOAD_FACTORY = 0.75f;

    LinkedHashMap<K, V> map;

    public SimpleLRUCache(int cacheSize) {
        MAX_CACHE_SIZE = cacheSize;
        int capacity = (int)Math.ceil(MAX_CACHE_SIZE / DEFAULT_LOAD_FACTORY) + 1;
        map = new LinkedHashMap<K, V>(capacity, DEFAULT_LOAD_FACTORY, true) {
            private static final long serialVersionUID = 1L;
            @Override
            protected boolean removeEldestEntry(Map.Entry<K, V> eldest) {
                return size() > MAX_CACHE_SIZE;
            }
        };
    }

    public synchronized void put(K key, V value) {
        map.put(key, value);
    }
    public synchronized V get(K key) {
        return map.get(key);
    }
    public synchronized void remove(K key) {
        map.remove(key);
    }
    public synchronized Set<Map.Entry<K, V>> getAll() {
        return map.entrySet();
    }
    @Override
    public String toString() {
        StringBuilder stringBuilder = new StringBuilder();
        for (Map.Entry<K, V> entry : map.entrySet()) {
            stringBuilder.append(String.format("%s: %s  ", entry.getKey(), entry.getValue()));
        }
        return stringBuilder.toString();
    }

    public static void main(String[] args) {
        SimpleLRUCache<String, String> lru = new SimpleLRUCache<>(5);
        lru.put("球员1", "杜兰特");
        lru.put("球员2", "表妹");
        lru.put("球员3", "库里");
        System.out.println(lru);
        lru.get("球员1");
        System.out.println(lru);
        lru.put("球员4", "一哥");
        lru.put("球员5", "汤姆");
        System.out.println(lru);
        lru.put("球员6", "格林");
        System.out.println(lru);
    }
}