Question

我正在尝试将两个（预先排序的）双向链接列表合并在一起，并且在尝试添加时会不断获得无限循环或一个元素列表。

以下代码的预期结果应该是列表：

[0,1,2,3,4,5,6,7,8,9]

所以我有：

public static void main(String[] args) {
    TheLinkedList<Integer> oddList = new TheLinkedList<Integer>();
    TheLinkedList<Integer> evenList = new TheLinkedList<Integer>();

    // Test lists 
    oddList.add(new Integer(9));
    oddList.add(new Integer(7));
    oddList.add(new Integer(5));
    oddList.add(new Integer(3));
    oddList.add(new Integer(2));
    oddList.add(new Integer(1));

    evenList.add(new Integer(8));
    evenList.add(new Integer(6));
    evenList.add(new Integer(4));
    evenList.add(new Integer(2));
    evenList.add(new Integer(0));

    //System.out.println(oddList.toString());
    //System.out.println(evenList.toString());
    oddList.merge(evenList);
    //System.out.println(theList.toString());
}

请注意，这是在另一个类中，您无法直接访问oddList或evenList

// Self explanatory getter and setter methods
public void add(T newValue) {
    head = new Node<T>(newValue, head, null);
    if (head.getNext() != null)
        head.getNext().setPrevious(head);
    else
        tail = head;
    count++;
}

public void merge(TheLinkedList<T> two) {
    do {

        if (head.getValue().compareTo(two.head.getValue()) <= 0) {
            head = head.getNext();
            continue;
        }
        if (head.getValue().compareTo(two.head.getValue()) >= 0){
            two.head = two.head.getNext();
        }
    } while (head != null && two.head != null);
}

Answer 1

我在这里看不到任何实际的合并？假设列表按降序排列，您应该将一个列表的头部与另一个列表的头部进行比较，如果另一个列表的值小于主列表的值，则应将该值插入主列表中到头节点的下一个节点，然后转到下一个值。您需要小心，因为添加到主列表中的节点需要引用下一个和上一个项目才能正确拟合。

如果订单无法保证，那么据我所知，您需要先对列表进行排序，然后将它们合并，以防止在最坏的情况下多次遍历整个引用的链表。

编辑：这是一个代码示例。你也可以通过递归来做到这一点，但是递归让我的头疼得那么......

public void merge(TheLinkedList<T> two) {
Node workingNodeOnOne = this.head;
Node workingNodeOnTwo = two.head;

while (workingNodeOnTwo != null)
        if (workingNodeOnOne.getValue().compareTo(workingNodeOnTwo.getValue()) < 0) {
            //this is if the value of your second working node is greater than the value of your first working node.
            //add the two.head.getValue() value of your current list here before the first working node...
            this.addBefore(workingNodeOnTwo.getValue(), workingNodeOnOne) 
            //note that this does change the value of this.head, but it doesn't matter as we are sorted desc anyways

            //given the constraints of what you have presented, you should never even hit this code block
            workingNodeOnTwo = workingNodeOnTwo.next();
        }
        else { //this is if the head value of second list is less than or equal to current head value, so just insert it after the current value here

            this.add(workingNodeOnTwo.getValue(), workingNodeOnOne); //insert the value of the second working node after our current working node
            workingNodeOnOne = workingNodeOnOne.next();
            workingNodeOnTwo = workingNodeOnTwo.next(); //go on to the next nodes
            }

}

这肯定需要一些调整，具体取决于你的实现，但逻辑是合理的，我相信。

Java - Doubly Linked List添加值

1 个答案: