public LinkedHashMap(int initialCapacity, float loadFactor) {
        super(initialCapacity, loadFactor);
        accessOrder = false;
    }
public LinkedHashMap(int initialCapacity) {
        super(initialCapacity);
        accessOrder = false;
    }
public LinkedHashMap() {
        super();
        accessOrder = false;
    }
public LinkedHashMap(Map<? extends K, ? extends V> m) {
        super();
        accessOrder = false;
        putMapEntries(m, false);
    }
public LinkedHashMap(int initialCapacity,
                         float loadFactor,
                         boolean accessOrder) {
        super(initialCapacity, loadFactor);
        this.accessOrder = accessOrder;
    }

/**
     * Returns <tt>true</tt> if this map should remove its eldest entry.
     * This method is invoked by <tt>put</tt> and <tt>putAll</tt> after
     * inserting a new entry into the map.  It provides the implementor
     * with the opportunity to remove the eldest entry each time a new one
     * is added.  This is useful if the map represents a cache: it allows
     * the map to reduce memory consumption by deleting stale entries.
     *
     * <p>Sample use: this override will allow the map to grow up to 100
     * entries and then delete the eldest entry each time a new entry is
     * added, maintaining a steady state of 100 entries.
     * <pre>
     *     private static final int MAX_ENTRIES = 100;
     *
     *     protected boolean removeEldestEntry(Map.Entry eldest) {
     *        return size() &gt; MAX_ENTRIES;
     *     }
     * </pre>
     *
     * <p>This method typically does not modify the map in any way,
     * instead allowing the map to modify itself as directed by its
     * return value.  It <i>is</i> permitted for this method to modify
     * the map directly, but if it does so, it <i>must</i> return
     * <tt>false</tt> (indicating that the map should not attempt any
     * further modification).  The effects of returning <tt>true</tt>
     * after modifying the map from within this method are unspecified.
     *
     * <p>This implementation merely returns <tt>false</tt> (so that this
     * map acts like a normal map - the eldest element is never removed).
     *
     * @param    eldest The least recently inserted entry in the map, or if
     *           this is an access-ordered map, the least recently accessed
     *           entry.  This is the entry that will be removed it this
     *           method returns <tt>true</tt>.  If the map was empty prior
     *           to the <tt>put</tt> or <tt>putAll</tt> invocation resulting
     *           in this invocation, this will be the entry that was just
     *           inserted; in other words, if the map contains a single
     *           entry, the eldest entry is also the newest.
     * @return   <tt>true</tt> if the eldest entry should be removed
     *           from the map; <tt>false</tt> if it should be retained.
     */
    protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {
        return false;
    }

public class Demo {
    static class MyMap<k,v> extends LinkedHashMap<k,v>{    //定义一个继承LinkedHashMap的类，并重写removeEldestEntry方法
        int size;
        public MyMap (int size){        //只提供一个有参的构造方法，用来设置map的大小
            /*
             * 这里调用的是LinkedHashMap的LinkedHashMap(int initialCapacity,boolean accessOrder)
             * 构造方法，设置按访问顺序排序
             */
            super(16,0.75f,true);        
            this.size = size;
        }
        @Override
        protected boolean removeEldestEntry(Entry<k, v> eldest) {
            // TODO Auto-generated method stub
            return this.size<size();    //设置当map里面存放的元素超过MyMap设置的size时就删除最老的元素
        }
        
    }
    public static void main(String[] args) {
        Map<String,String> map = new MyMap<String,String>(5);    
        for (int i = 0; i < 10; i++) {
            map.put(i+"", i+"");
            System.out.print(i+"="+map.size()+"\t");
        }
        System.out.println();
        //遍历map看看里面的元素
        map.forEach((k,v)->{
            System.out.println("k="+k);
        });
        
        //此时访问一下里面的元素，再观察遍历后的顺序
        map.get("6");map.get("8");
        System.out.println("========================");
        map.forEach((k,v)->{
            System.out.println("k="+k);
        });
    }
}

打印的结果：
　　

0=1 1=2 2=3 3=4 4=5 5=5 6=5 7=5 8=5 9=5
k=5
k=6
k=7
k=8
k=9
========================
k=5
k=7
k=9
k=6
k=8