Question

为了尝试在Java中实现Banker的算法，我试图迭代存储多个线程的id的ArrayList。我想继续迭代这个数组，直到找到一个安全的序列（或不是）。到目前为止，我只能遍历ArrayList一次，但是如果我发现它们是我安全序列中的下一个，我需要将对象删除。

我如何以循环方式迭代遍历ArrayList（或任何其他数据结构），因为我去的时候从中删除对象？我希望我的问题很清楚，但到目前为止我的内容是：在最好的情况下，当列表中剩余0个对象时，算法停止。

修改我的意思是循环时尚是这样的：0,1,2,3,4,0,1,2,3,4（我还需要能够从列表中删除进程，因此迭代应该变小。）可能看起来像：注意当剩余列表中有0时，算法会停止

0,1,2,3,4

，0,2,3,4

2,3,4

2,4

4

while (it.hasNext()) {
            Integer thisOne = it.next();
            found=false;
            // this process, can be removed from the list
            // because cN = its current need
            // a = how much the process is allocated already
            // cR=bank's current resources
            // so if cN - a <= cR, we know there is no deficit
            // we can satisfy this process, and remove it from the list 
            // of needy processes

            if(noDeficit(cN[thisOne], a[thisOne], cR)){

                // we take away its resources and give them to the bank
                Utility.add(cR, a[thisOne]);
                Utility.zeroOut(cN[thisOne]);
                Utility.zeroOut(a[thisOne]);

                found = true;
                // we remove it from the list of processes 
                it.remove();                                    
            }if(!found) return false;

        }

Answer 1

没有真正得到你的“发现”条件应该做的事情，但这是一种方法。

    //todo  create loop condition for stopping, eg originalList.size() > 0
    //eg
    List<Integer> itemsToRemoveList = new ArrayList<Integer>();
    while (originalList.size() > 0) { 
       Iterator it = originalList.iterator();      

       while (it.hasNext()) {
          Integer thisOne = it.next();
          found=false;
          // this process, can be removed from the list
          if(noDeficit(cN[thisOne], a[thisOne], cR)){

              // we take away its resources and give them to the bank
              Utility.add(cR, a[thisOne]);
              Utility.zeroOut(cN[thisOne]);
              Utility.zeroOut(a[thisOne]);

              // we remove it from the list of processes 
              itemsToRemoveList.add(thisOne);
           }                                   
        } //end loop

        for (Integer integerToRemove : itemsToRemoveList) {  
           originalList.remove(integerToRemove);
        }
        itemsToRemoveList.clear();

   }

如何从中删除对象时继续迭代ArrayList

1 个答案: