Question

我正在开展一个项目，我需要使用递归，以及arraylists / vector来解决3-satisfiability实例。任务是选择整数子集，使得下面表示的每个集合包含子集的一个或多个元素。约束是一个数字，i＆amp;相反，-i，不能同时在子集中。

Vector<Vector<Integer>> matrix = new Vector<Vector<Integer>>();
for(int i=0;i<4;i++)
{
    matrix.add(new Vector<Integer>());
}

然后用数字填充二维向量/ arraylist，例如：

1, -2, 3
-1, 2, 4
2, -3, 4
-1, -2, -3

if(startRec(matrix)
{
   //solution is found
}
else
{ 
   //no solution is possible
}


private boolean startRec(Vector<Vector<Integer>> matrix)
{  // I'm using trial and error to find a solution to the above problem.  So in the below, matrix.get(0).get(i) is selected as part of a possibly correct set.  
    boolean success=false;
    if(stagen(matrix.get(0).get(0), matrix))
    {
        success=true;
    }
    else if(stagen(matrix.get(0).get(1),matrix))
    {
        success=true;
    }
    else if(stagen(matrix.get(0).get(2),matrix))
    {
        success=true;
    }

    return success;
}
private boolean stagen(int input, Vector<Vector<Integer>> matrix)
{
   removei(input, matrix) // this removes all Vector<Integer> that contain i.  Those sets are considered satisfied, and no longer need to be addressed.
   removenegative(input,matrix) // this removes all integers that are -input.  Since i and -i cannot both be selected, I'm removing the -input. 
//So if a set contained three ints, one of which was -input, it now contains 2.
    boolean success=false;
    if(stagen(matrix.get(0).get(0), matrix)) // since the original matrix.get(0) contained input, it was removed in removei(input,matrix), thus this is a one below the one called previously.
    {
        success=true;
    }
    else if(stagen(matrix.get(0).get(1),matrix))
    {
        success=true;
    }
    else if(stagen(matrix.get(0).get(2),matrix))
    {
        success=true;
    }

    return success;
}

我已删除超出范围的检查，以使其更具可读性。但我充满信心的过程就是：

 1, -2, 3   //1 is selected in the second line of startRec
-1, 2, 4
2, -3, 4
-1, -2, -3

 //1, -2, 3 line is removed from consideration by method removei() as 1 has been selected
 2, 4       // -1 has been removed as both 1,-1 cannot be selected.
2, -3, 4
-2, -3     // -1 removed.

 2, 4       now 2 is the first number, so 2 is selected.
-2, 3, 4   
-2, -3 

//2, 4     removed via removei method.
3, 4       //-2 is removed, because 2 has been selected.
-3         //-2removed.

3, 4    //Now 3 is selected.
-3

//3, 4  line removed as it has been satisfied.
_____   //There's now an empty set here as -3 was deleted 
//false is returned by stagen.

程序返回到选择3的stagen，现在理想情况下现在选择4作为下一个字母。但我删除了整行。

有没有办法在不创建未知数量的向量的情况下执行此操作？

2，arraylists＆amp; amp;引导？我一直在使用arraylists，但项目建议使用矢量。

Answer 1

你问题的第一部分是不连贯的。但是，我将详细阐述Vector和Arraylist之间的区别：

如果多个线程同时访问ArrayList，则需要外部同步修改列表的代码块。
在内部，它们都使用数组来维护数据结构。如果空间不足，ArrayList会将其大小增加50％，而Vector会将其大小增加一倍。

请参阅this主题以获取详细说明。

ArrayLists，References，＆amp;递归

1 个答案: