【问题标题】:Java recursive method end but doesn't terminateJava递归方法结束但不终止
【发布时间】:2013-10-29 03:39:42
【问题描述】:

我正在为双链表数据结构编写一个方法,该方法应该按照参数中的指定将元素列表从索引 a 反转到 b,并决定递归地执行此操作。作业的目的是使用节点练习指针操作。我的方法的逻辑对我来说似乎没问题,但是当我通过 JUnit 测试运行代码时,它不会结束。觉得这很奇怪,我添加了 println 语句来查看它到达了代码的哪些部分;一切都检查好了。所以我通过 Eclipse 的调试器运行它,它到达了结束括号,在向前和向后遍历所有递归调用之后,没有终止。它只是坐在末端支架上,我以前从未见过这样的东西。为什么会这样?我能做些什么来解决它?

代码如下:

public void reverseList(int start, int end)
{
    if (start >= end)
    {
        return;
    }

    ListNode left = getListNode(start);
    ListNode right = getListNode(end);
    ListNode leftNext = left.next;
    ListNode leftPrev = left.previous;
    ListNode rightNext = right.next;
    ListNode rightPrev = right.previous;

    leftPrev.next = right;
    rightPrev.next = left;
    leftNext.previous = right;
    rightNext.previous = left;
    left.next = rightNext;
    left.previous = rightPrev;
    right.next = leftNext;
    right.previous = leftPrev;

    reverseList(start + 1, end - 1);
}

编辑:这是测试它的代码

JUnit:

@Test
public void testReveseList()
{
StudentList list = new StudentList();
list.add("a", "");
list.add("b", "");
list.add("c", "");
list.add("d", "");
list.add("e", "");
list.add("f", "");
list.add("g", "");
list.add("h", "");
list.add("i", "");
list.add("j", "");
list.printlist();
list.reverseList(2, 5);
System.out.println();
StudentList expectedList = new StudentList();
expectedList.add("a", "");
expectedList.add("b", "");
expectedList.add("f", "");
expectedList.add("e", "");
expectedList.add("d", "");
expectedList.add("c", "");
expectedList.add("g", "");
expectedList.add("h", "");
expectedList.add("i", "");
expectedList.add("j", "");
assertEquals(expectedList, list);
list.reverseList(2, 5);

System.out.println();

StudentList expectedList1 = new StudentList();
expectedList1.add("a", "");
expectedList1.add("b", "");
expectedList1.add("c", "");
expectedList1.add("d", "");
expectedList1.add("e", "");
expectedList1.add("f", "");
expectedList1.add("g", "");
expectedList1.add("h", "");
expectedList1.add("i", "");
expectedList1.add("j", "");
assertEquals(expectedList1, list);
list.reverseList(0, 9);
System.out.println();
StudentList expectedList2 = new StudentList();
expectedList2.add("j", "");
expectedList2.add("i", "");
expectedList2.add("h", "");
expectedList2.add("g", "");
expectedList2.add("f", "");
expectedList2.add("e", "");
expectedList2.add("d", "");
expectedList2.add("c", "");
expectedList2.add("b", "");
expectedList2.add("a", "");
assertEquals(expectedList2, list);

}

测试类:

public class StudentList
{
private ListNode head = null;

public void add(StudentData data)
{
    ListNode newNode = new ListNode(data);
    ListNode lastNode = getTail();

    if (lastNode == null)
    {
        head = newNode;
    }
    else
    {
        lastNode.next = newNode;
    }

    newNode.previous = lastNode;
    newNode.next = null;

}       

    public ListNode getListNode(int indexOfDesiredNode)
{
    if (indexOfDesiredNode >= size())
    {
        // --- Error: There aren't that many nodes
        return null;
    }

    // --- Move through the list, node by node, until we find
    // --- the one we want
    int count = 0;
    ListNode current = head;
    while ((count < indexOfDesiredNode) && (null != current))
    {
        current = current.next;
        count = count + 1;
    }

    // --- So, did we find anything?
    if (null == current)
    {
        // --- Error: We didn't find the node
        return null;
    }

    return current;
}

    public int size()
{
    if (null == head)
    {
        return 0;
    }

    int count = 0;
    ListNode current = head;
    while (current != null)
    {
        count = count + 1;
        current = current.next;
    }
    return count;
}

    public void reverseList(int start, int end)
{
    if (start >= end)
    {
        return;
    }

    ListNode left = getListNode(start);
    ListNode right = getListNode(end);
    ListNode leftNext = left.next;
    ListNode leftPrev = left.previous;
    ListNode rightNext = right.next;
    ListNode rightPrev = right.previous;

    leftPrev.next = right;
    rightPrev.next = left;
    leftNext.previous = right;
    rightNext.previous = left;
    left.next = rightNext;
    left.previous = rightPrev;
    right.next = leftNext;
    right.previous = leftPrev;

    //-- The easy way of doing it... *sigh*
    // StudentData temp = get(start);
    // getListNode(start).data = getListNode(end).data;
    // getListNode(end).data = temp;

    reverseList(start + 1, end - 1);
}
}

和 ListNode 类:

public class ListNode
{
public StudentData data = null;
public ListNode next = null;
public ListNode previous = null;

public ListNode(StudentData data)
{
    this.data = data;
}

@Override
public String toString()
{
    return data.toString();
}
}

很抱歉,格式不是最好的,复制和粘贴时发生了这种情况。

【问题讨论】:

  • 它应该可以工作。问题可能出在getListNode 电话中。请问可以发SSCCE吗?
  • 这是一个尾递归函数;因此,在最后一个递归调用实际返回后,您只需通过到达函数末尾来隐式地从每个连续调用中返回。你确定它永远不会终止吗?
  • getListNode 是我的导师写的,所以我很肯定它有效。它只是返回指定索引处的节点。
  • @user2900718 那么你能发布一个工作示例吗?假设getListNode 方法正常工作,您的代码就可以正常工作。
  • 在eclipse中,它不会显示println语句无限打印,但我必须手动终止程序。

标签: java eclipse list recursion linked-list


【解决方案1】:

问题在于end == start + 1,因此rightPrev == left。然后你设置rightPrev.next = left 使你的列表循环,所以getListNode 不会终止。

【讨论】:

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