我正在一个项目中，我有十个文件，每个文件我必须使用贪婪算法10次。但是，文件具有1000到9000个元素的巨大输入。下面的代码仅是一个包含1000个元素的文件，我运行了10次。我计算平均运行时间和其他2个值，仅一个文件就需要15分钟。通过哪些更改可以使过程更快？ （所有代码运行的方法完全没有问题）

    public static void main(String[] args) {


    long[] avg1=new long[10]; //stores running time of each try in array 
    long sum=0; //stores average running time 
    int gval=0;
    int optval=0;
    for(int i =0;i<10;i++)
    {   
        BuyersList b = new BuyersList();
        BuyersList buyers = new BuyersList();
        buyers.readFile("src/p500x1000.txt"); //reads file and adds values to list 
        long start = System.currentTimeMillis();
        b=buyers.greedy(500); //algorithm runs 
        long end = System.currentTimeMillis();    
        avg1[i]=(end-start); //calculate running time of greedy algorithm
        sum+=avg1[i];  //add time of one try to sum
        gval=b.totalValue();
        optval=buyers.opt;  
    }


    System.out.println("m= 500 :");    // for 500 objects print the values below 
    System.out.println(" n = 1000 : avgTime = " + sum/10 + " greedy value = " + gval + " opt value = " + optval + " ");



    }

贪婪算法：

public BuyersList greedy( int m  ) {

    if(empty())
    {
        return null;
    }
    ItemsList lista = new ItemsList(); // create list of items 
    BuyerNode current=this.first;     // node that represents first buyer 
    boolean f = true;   //suppose one subset list exists 

    for(int i=0;i<m;i++)  //enter pointers of items 
    {
        lista.append(i);
    }

    BuyersList best = new BuyersList(); //create the returning buyers list 


    while(f==true)    //while one subset list of complete list  exists in the list of buyers 
    {
            current=this.maximum(this); //call maximum function to set current to maximum fraction 
            boolean flag = lista.contains(current.itemsList);   // if current's list is a subset 
            if(flag==true){
            best.append(current.id,current.value,current.itemsList); //append current to the best list 
            lista.remove(current.itemsList); //remove current's list  from ItemsList 
            this.deleteNode(current.id); // call private deletenode function to delete current from buyers
            this.nbNodes--;
          }

            else {this.deleteNode(current.id);}


        BuyerNode bnode = this.first;
        f = false;  //set f to false to exit loop if it doesn't change below 
        for(bnode=this.first;bnode!=null;bnode=bnode.next)
        {   

            if(lista.contains(bnode.itemsList))  //check if itemslist contains even one buyer subset list 
            {               

                f = true;   //one subset list exists so loop again 
            }
        }

    } 




    return best;
}